请问：y=a1.*exp(-excel sqrt函数(...
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请问：y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))printf()函数中’\n’；’\t’；’\a_百度知道
请问：y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))printf()函数中’\n’；’\t’；’\a
B我们讨论的范围是x&0\AD=AB BD=AB BC/2
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x-1BE=BC CE=BC CA/0所以f(x)=1
b f(x)=x-1
g(x)=x^2/0所以|x-1| |x-2|>mx2-(1-m)x 1>
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好吧，我承认我标题党了，不过既然你来了，就认真看下去吧，保证你有收获。
我们平时经常会有一些数据运算的操作，需要调用sqrt，exp，abs等函数，那么时候你有没有想过：这个些函数系统是如何实现的？就拿最常用的sqrt函数来说吧，系统怎么来实现这个经常调用的函数呢？
虽然有可能你平时没有想过这个问题，不过正所谓是&临阵磨枪，不快也光&，你&眉头一皱，计上心来&，这个不是太简单了嘛，用二分的方法，在一个区间中，每次拿中间数的平方来试验，如果大了，就再试左区间的中间数；如果小了，就再拿右区间的中间数来试。比如求sqrt(16)的结果，你先试(0+16)/2=8，8*8=64，64比16大，然后就向左移，试(0+8)/2=4，4*4=16刚好，你得到了正确的结果sqrt(16)=4。然后你三下五除二就把程序写出来了：
float SqrtByBisection(float n) //用二分法
if(n&0) //小于0的按照你需要的处理
float mid,
float low,
low=0, up=n;
mid=(low+up)/2;
if(mid*mid&n)
mid=(up+low)/2;
}while(abs(mid-last) & eps);//精度控制
然后看看和系统函数性能和精度的差别（其中时间单位不是秒也不是毫秒，而是CPU Tick，不管单位是什么，统一了就有可比性）&
从图中可以看出，二分法和系统的方法结果上完全相同，但是性能上整整差了几百倍。为什么会有这么大的区别呢？难道系统有什么更好的办法？难道。。。。哦，对了，回忆下我们曾经的高数课，曾经老师教过我们&牛顿迭代法快速寻找平方根&，或者这种方法可以帮助我们，具体步骤如下：
求出根号a的近似值：首先随便猜一个近似值x，然后不断令x等于x和a/x的平均数，迭代个六七次后x的值就已经相当精确了。 例如，我想求根号2等于多少。假如我猜测的结果为4，虽然错的离谱，但你可以看到使用牛顿迭代法后这个值很快就趋近于根号2了： (4 + 2/4) / 2 = 2.25 (2.25 + 2/2.25) / 2 = 1.56944.. (1.56944.. + 2/1.56944..) / 2 = 1.42189.. (1.42189.. + 2/1.42189..) / 2 = 1.41423.. ....
这种算法的原理很简单，我们仅仅是不断用(x,f(x))的切线来逼近方程x^2-a=0的根。根号a实际上就是x^2-a=0的一个正实根，这个函数的导数是2x。也就是说，函数上任一点(x,f(x))处的切线斜率是2x。那么，x-f(x)/(2x)就是一个比x更接近的近似值。代入 f(x)=x^2-a得到x-(x^2-a)/(2x)，也就是(x+a/x)/2。
相关的代码如下：
float SqrtByNewton(float x)
float val = //最终
float //保存上一个计算的值
val =(val + x/val) / 2;
}while(abs(val-last) & eps);
然后我们再来看下性能测试：
哇塞，性能提高了很多，可是和系统函数相比，还是有这么大差距，这是为什么呀？想啊想啊，想了很久仍然百思不得其解。突然有一天，我在网上看到一个神奇的方法，于是就有了今天的这篇文章，废话不多说，看代码先：
float InvSqrt(float x)
float xhalf = 0.5f*x;
int i = *(int*)&x; // get bits for floating VALUE
i = 0x5f375a86- (i&&1); // gives initial guess y0
x = *(float*)&i; // convert bits BACK to float
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
return 1/x;
然后我们最后一次来看下性能测试：
这次真的是质变了，结果竟然比系统的还要好。。。哥真的是震惊了！！！哥吐血了！！！一个函数引发了血案！！！血案，血案。。。
到现在你是不是还不明白那个&鬼函数&，到底为什么速度那么快吗？不急，先看看下面的故事吧：
Quake-III Arena (雷神之锤3)是90年代的经典游戏之一。该系列的游戏不但画面和内容不错，而且即使计算机配置低，也能极其流畅地运行。这要归功于它3D引擎的开发者约翰&卡马克（John Carmack）。事实上早在90年代初DOS时代，只要能在PC上搞个小动画都能让人惊叹一番的时候，John Carmack就推出了石破天惊的Castle Wolfstein，然后再接再励，doom, doomII, Quake...每次都把3-D技术推到极致。他的3D引擎代码资极度高效，几乎是在压榨PC机的每条运算指令。当初MS的Direct3D也得听取他的意见，修改了不少API。
最近，QUAKE的开发商ID SOFTWARE 遵守GPL协议，公开了QUAKE-III的原代码，让世人有幸目睹Carmack传奇的3D引擎的原码。这是QUAKE-III原代码的下载地址：
(下面是官方的下载网址，搜索 &quake3-1.32b-source.zip& 可以找到一大堆中文网页的。
我们知道，越底层的函数，调用越频繁。3D引擎归根到底还是数学运算。那么找到最底层的数学运算函数（在game/code/q_math.c）， 必然是精心编写的。里面有很多有趣的函数，很多都令人惊奇，估计我们几年时间都学不完。在game/code/q_math.c里发现了这样一段代码。它的作用是将一个数开平方并取倒，经测试这段代码比(float)(1.0/sqrt(x))快4倍：
float Q_rsqrt( float number )
const float threehalfs = 1.5F;
x2 = number * 0.5F;
= * ( long * ) &y;
// evil floating point bit level hacking
= 0x5f3759df - ( i && 1 ); // what the fuck?
= * ( float * ) &i;
= y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
= y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
#ifndef Q3_VM
#ifdef __linux__
assert( !isnan(y) ); // bk010122 - FPE?
函数返回1/sqrt(x)，这个函数在图像处理中比sqrt(x)更有用。&
注意到这个函数只用了一次叠代！（其实就是根本没用叠代，直接运算）。编译，实验，这个函数不仅工作的很好，而且比标准的sqrt()函数快4倍！要知道，编译器自带的函数，可是经过严格仔细的汇编优化的啊！
这个简洁的函数，最核心，也是最让人费解的，就是标注了&what the fuck?&的一句：
i = 0x5f3759df - ( i && 1 );
= y * ( threehalfs - ( x2 * y * y ) );
两句话就完成了开方运算！而且注意到，核心那句是定点移位运算，速度极快！特别在很多没有乘法指令的RISC结构CPU上，这样做是极其高效的。
算法的原理其实不复杂，就是牛顿迭代法，用x-f(x)/f'(x)来不断的逼近f(x)=a的根。
没错，一般的求平方根都是这么循环迭代算的，但是卡马克(quake3作者)真正牛B的地方是他选择了一个神秘的常数 0x5f3759df 来计算那个猜测值，就是我们加注释的那一行，那一行算出的值非常接近1/sqrt(n)，这样我们只需要2次牛顿迭代就可以达到我们所需要的精度。好吧如果这个还不算NB，接着看:
普渡大学的数学家Chris Lomont看了以后觉得有趣，决定要研究一下卡马克弄出来的这个猜测值有什么奥秘。Lomont也是个牛人，在精心研究之后从理论上也推导出一个最佳猜测值，和卡马克的数字非常接近，0x5f37642f。卡马克真牛，他是外星人吗？
传奇并没有在这里结束。Lomont计算出结果以后非常满意，于是拿自己计算出的起始值和卡马克的神秘数字做比赛，看看谁的数字能够更快更精确的求得平方根。结果是卡马克赢了... 谁也不知道卡马克是怎么找到这个数字的。
最后Lomont怒了，采用暴力方法一个数字一个数字试过来，终于找到一个比卡马克数字要好上那么一丁点的数字，虽然实际上这两个数字所产生的结果非常近似，这个暴力得出的数字是0x5f375a86。
Lomont为此写下一篇论文，"Fast Inverse Square Root"。 论文下载地址：
参考：&IEEE Standard 754 for Binary Floating-Point Arithmetic&&FAST INVERSE SQUARE ROOT&
最后，给出最精简的1/sqrt()函数：
float InvSqrt(float x)
float xhalf = 0.5f*x;
int i = *(int*)&x; // get bits for floating VALUE
i = 0x5f375a86- (i&&1); // gives initial guess y0
x = *(float*)&i; // convert bits BACK to float
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
大家可以尝试在PC机、51、AVR、430、ARM、上面编译并实验，惊讶一下它的工作效率。
前两天有一则新闻，大意是说 Ryszard Sommefeldt 很久以前看到这么样的一段 code (可能出自 Quake III 的 source code)：
float InvSqrt (float x)
float xhalf = 0.5f*x;
int i = *(int*)&x;
i = 0x5f3759df - (i&&1);
x = *(float*)&i;
x = x*(1.5f - xhalf*x*x);
他一看之下惊为天人，想要拜见这位前辈高人，但是一路追寻下去却一直找不到人；同时间也有其他人在找，虽然也没找到出处，但是 Chris Lomont 写了一篇论文 (in PDF) 解析这段 code 的算法 (用的是 Newton&s Method，牛顿法；比较重要的是后半段讲到怎么找出神奇的 0x5f3759df 的)。&
PS. 这个 function 之所以重要，是因为求&开根号倒数&这个动作在 3D 运算 (向量运算的部份) 里面常常会用到，如果你用最原始的 sqrt() 然后再倒数的话，速度比上面的这个版本大概慢了四倍吧& XD PS2. 在他们追寻的过程中，有人提到一份叫做 MIT HACKMEM 的文件，这是 1970 年代的 MIT 强者们做的一些笔记 (hack memo)，大部份是 algorithm，有些 code 是 PDP-10 asm 写的，另外有少数是 C code (有人整理了一份列表)
好了，故事就到这里结束了，希望大家能有有收获:)
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编程基础最新文章请问一下y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))AB=AC=3COS=1/9_百度知道
请问一下y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))AB=AC=3COS=1/9
oga[(x2^2-ax2)/0f(x)=x^5/5-ax^3/(x1^2-ax1)]>
提问者采纳
2<,0)-1/2;x1<AB BC CA)/x2&2=0f(x)=loga(x^2-ax)在(-1/0仿照线相交于点O仿照AB BC CA)/
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Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
(Formerly M)
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, , , 1, 2, 7, , 49, 05, 088169
Also sometimes called Lamé's sequence.
F(n+2) = number of binary sequences of length n that have no consecutive 0's.
F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.
F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.
F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - , Jun 18 2001
F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]
Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y &= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z&0 then z=F(n + 1).
For Fibonacci search see Knuth, Vol. 3; Horowitz and S etc.
F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - , Dec 29 2001
F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(), May 19 2002
F(n+1) = number of ()-, ()- and ()-avoiding involutions in S_n.
This is also the Horadam sequence (0,1,1,1). - , Aug 18 2003
An INVERT transform of . INVERT([1,1,2,3,5,8,...]) gives . INVERT([1,2,3,5,8,13,21,...]) gives . - , Dec 12 2003
Number of meaningful differential operations of the k-th order on the space R^3. - , Mar 02 2004
F(n)=number of compositions of n-1 with no part greater than 2. Example: F(4)=3 because we have 3 = 1+1+1 = 1+2 = 2+1.
F(n) = number of compositions e.g., F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - , Jun 22 2004
F(n) = number of binary words of length n beginning with 0 and havin e.g., F(6) counts , , 0001. - , Jun 22 2004
The number of sequences (s(0),s(1),...,s(n)) such that 0&s(i)&5, |s(i)-s(i-1)|=1 and s(0)=1 is F(n+1); e.g., F(5+1) = 8 corresponds to , , 3434. - , Jun 22 2004 [corrected by Neven Juric, Jan 09 2009]
Likewise F(6+1) = 13 corresponds to these thirteen sequences with seven numbers: 11, 11, 33, 33, 1234343. - Neven Juric, Jan 09 2008
A relationship between F(n) and the Mandelbrot set is discussed in the link &Le nombre d'or dans l'ensemble de Mandelbrot& (in French). - , Sep 19 2004
For n&0, the continued fraction for F(2n-1)*Phi=[F(2n);L(2n-1),L(2n-1),L(2n-1),...] and the continued fraction for F(2n)*Phi=[F(2n+1)-1;1,L(2n)-2,1,L(2n)-2,...]. Also true: F(2n)*Phi=[F(2n+1);-L(2n),L(2n),-L(2n),L(2n),...] where L(i) is the i-th Lucas number ().... - , Nov 28 2004 [corrected by , Oct 20 2010]
F(n+1) (for n&=1) = number of permutations p of 1,2,3,...,n such that |k-p(k)|&=1 for k=1,2,...,n. (For &=2 and &=3, see
and .) - , Nov 28 2004
The ratios F(n+1)/F(n) for n&0 are the convergents to the simple continued fraction expansion of the golden section. - , Dec 19 2004
Lengths of successive words (starting with a) under the substitution: {a -& ab, b -& a}. - , Jan 22 2005
The Fibonacci sequence, like any additive sequence, naturally tends to be geometric with common ratio not a rational power of 10; consequently, for a sufficiently large number of terms, Benford's law of first significant digit (i.e., first digit 1 &= d &= 9 occurring with probability log_10(d+1) - log_10(d)) holds. - , Apr 29 2005
a(n) = Sum(abs((n, k)): 0 &= k &= n). - , Jun 01 2005
a(n) = ((n)).
Fib(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of . - , Mar 11 2003
Number of order ideals of the &zig-zag& poset. See vol. 1, ch. 3, prob. 23 of Stanley. - , Dec 27 2005
F(n+1)/F(n) is also the Farey fraction sequence (see
for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - , May 08 2006
a(n+2) is the number of paths through 2 plates of glass with n reflections (reflections occurring at plate/plate or plate/air interfaces). Cf. -. - , Jul 06 2006
F(n+1) equals the number of downsets (i.e., decreasing subsets) of an n-element fence, i.e., an ordered set of height 1 on {1,2,...,n} with 1 & 2 & 3 & 4 & ... n and no other comparabilities. Alternatively, F(n+1) equals the number of subsets A of {1,2,...,n} with the property that, if an odd k is in A, then the adjacent elements of {1,2,...,n} belong to A, i.e., both k - 1 and k + 1 are in A (provided they are in {1,2,...,n}). - , Aug 25 2006
Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - , Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, round(log_phi(sqrt((sqrt(5) a(n) + sqrt(5 a(n)^2 - 4))(sqrt(5) a(n) + sqrt(5 a(n)^2 + 4)))/2)) = n for n &= 3, obtained by rounding the arithmetic mean of the inverses given in
and . - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
A result of Jacobi from 1848 states that every symmetric matrix over a p.i.d. is congruent to a triple-diagonal matrix. Consider the maximal number T(n) of summands in the determinant of an n X n triple-diagonal matrix. This is the same as the number of summands in such a determinant in which the main-, sub- and super-diagonal elements are all nonzero. By expanding on the first row we see that the sequence of T(n)'s is the Fibonacci sequence without the initial stammer on the 1's. - Larry Gerstein (gerstein(AT)math.ucsb.edu), Mar 30 2007
Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd. There is a similar representation for Lucas numbers (). Many Fibonacci formulas now easily follow from appropriate sinh and cosh formulas. For example: the de Moivre theorem (cosh(x)+sinh(x))^m=cosh(mx)+sinh(mx) produces L(n)^2+5F(n)^2=2L(2n) and L(n)F(n)=F(2n) (setting x=n*psi and m=2). - , Apr 18 2007
Inverse: floor(log_phi(sqrt(5)*Fib(n))+1/2)=n, for n&1. Also for n&0, floor(1/2*log_phi(5*Fib(n)*Fib(n+1)))=n. Extension valid for integer n, except n=0,-1: floor(1/2*sign(Fib(n)*Fib(n+1))*log_phi|5*Fib(n)*Fib(n+1)|)=n (where sign(x) = sign of x). - , May 02 2007
F(n+2) = The number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007
From Kauffman and Lopes, Proposition 8.2, p. 21: &The sequence of the determinants of the Fibonacci sequence of rational knots is the Fibonacci sequence (of numbers).& - , Oct 26 2007
This is a_1(n) in the Doroslovacki reference.
Let phi = (sqrt(5)+1)/2 = 1.6180339...; then phi^n = (1/phi)*a(n) + a(n+1). Example: phi^4 = 6.8541019... = (0.6180339...)*3 + 5. Also phi = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + 1/(34*89) + ... - , Dec 15 2007
The sequence of first differences, Fib(n+1)-Fib(n), is essentially the same sequence: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... - , Mar 03 2008
a(n)= the number of different ways to run up a staircase with n steps, taking steps of odd sizes where the order is relevant and there is no other restriction on the number or the size of each step taken. - , May 21 2008
Equals row sums of triangle . - , Sep 12 2008
Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - , Sep 15 2008
F(n) is the number of possible binary sequences of length n that obey the sequential construction rule: if last symbol is 0, add the complement (1); else add 0 or 1. Here 0,1 are metasymbols for any 2-valued symbol set. This rule has obvious similarities to JFJ Laros's rule, but is based on addition rather than substitution and creates a tree rather than a single sequence. - , Oct 05 2008
F(n) = Product_{k=1..(n-1)/2} (1 + 4*cos^2 k*Pi/n), where terms = roots to the Fibonacci product polynomials, . - , Nov 22 2008
Fp == 5^((p-1)/2) mod p, p = prime [Schroeder, p. 90]. -
& , Feb 21 2009
(Ln)^2 - 5*(Fn)^2 = 4*(-1)^n. Example: 11^2 - 5*5 = -4. - , Mar 11 2009
Output of Kasteleyn's formula for the number of perfect matchings of an m X n grid specializes to the Fibonacci sequence for m=2. - Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009
(Fib(n),Fib(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 06 2009
(Fib(n),Fib(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 08 2009
a(n+2)=((n)). - , Sep 26 2009
Difference between of number of closed walks of length n+1 from a node on a pentagon and number of walks of length n+1 between two adjacent nodes on a pentagon. - , Feb 10 2010
F(n+1) = number of Motzkin paths of length n having exactly one weak ascent. A Motzkin path of length n is a lattice path from (0,0) to (n,0) consisting of U=(1,1), D=(1,-1) and H=(1,0) steps and never going below the x-axis. A weak ascent in a Motzkin path is a maximal sequence of consecutive U and H steps. Example: a(5)=5 because we have (HHHH), (HHU)D, (HUH)D, (UHH)D, and (UU)DD (the unique weak ascent is shown see ). - , Mar 11 2010
(F(n-1) + F(n+1))^2 - 5F(n-2)*F(n+2) = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Mar 31 2010
From the Pinter and Ziegler reference's abstract: authors &show that essentially the Fibonacci sequence is the unique binary recurrence which contains infinitely many three-term arithmetic progressions. A criterion for general linear recurrences having infinitely many three-term arithmetic progressions is also given.& - , May 22 2010
F(n+1) = number of paths of length n starting at initial node on the path graph P_4. - , May 27 2010
F(k) = Number of cyclotomic polynomials in denominator of generating function for number of ways to place k nonattacking queens on an n X n board. - , Jun 07 2010
As n-& inf., (a(n)/a(n-1) - a(n-1)/a(n)) tends to 1.0. Example: a(12)/a(11) - a(11)/a(12) = 144/89 - 89/144 = 0..... - , Jul 16 2010
From , Oct 20 2010: (Start)
Fibonacci numbers are those numbers m such that m*phi is closer to an integer than k*phi for all k, 1&=k&m. More formally: a(0)=0, a(1)=1, a(2)=1, a(n+1)=minimal m&a(n) such that m*phi is closer to an integer than a(n)*phi.
For all numbers 1&=k&Fib(n), the inequality |k*phi-round(k*phi)| & |Fib(n)*phi-round(Fib(n)*phi)| holds.
Fib(n)*phi - round(Fib(n)*phi) = -((-phi)^(-n)), for n&1.
Fract(0.5+Fib(n)*phi) = 0.5 -(-phi)^(-n), for n&1.
Fract(Fib(n)*phi) = (1/2)*(1+(-1)^n)-(-phi)^(-n), n&1.
Inverse: n = -log_phi |0.5-fract(0.5+Fib(n)*phi)|.
F((n)*k) mod n = 0, for any integer k. - , Nov 27 2010
F(n+k)^2-F(n)^2 = F(k)*F(2n+k), for even k. - , Dec 04 2010
F(n+k)^2+F(n)^2 = F(k)*F(2n+k), for odd k. - , Dec 04 2010
F(n) = round(phi* F(n-1)) for n&1. - , Jan 13 2012
For n & 0: a(n) = length of n-th row in Wythoff array . - , Jan 26 2012
From , Feb 22 2012: (Start)
The number of free permutations of [n].
The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).
The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)
The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - , Apr 19 2012
Two conjectures: For n & 1, F(n+2)^2 mod F(n+1)^2 = F(n)*F(n+1) - (-1)^n. For n & 0, (F(2n) + F(2n+2))^2 = F(4n+3) + sum_{k = 2..2n}F(2k). - , May 06 2012
From , Jan 30 2014: (Start)
Proof of Ratushnyak's first conjecture: For n & 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2F(n+1)^2.
Consider: F(n+2)^2 - F(n)*F(n+1) - 2F(n+1)^2
&&&&&&&&&= F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)
&&&&&&&&&=(F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))
&&&&&&&&&= F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.
Proof of second conjecture: L(n) stands for Lucas number sequence from .
Consider the fact that
&&&&L(2n+1)^2 = L(4n+2) - 2
&&&(F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2
&&&(F(2n) + F(2n+2))^2 = sum{k = 2..2n, F(2k)} + F(4n+3).
The relationship: INVERT transform of (1,1,0,0,0,...) = (1, 2, 3, 5, 8,...), while the INVERT transform of (1,0,1,0,1,0,1,...) = (1, 1, 2, 3, 5, 8,...) is equivalent to: The numbers of compositions using parts 1 and 2 is equivalent to the numbers of compositions using parts == 1 mod 2 (i.e., the odd integers). Generally, the numbers of compositions using parts 1 and k is equivalent to the numbers of compositions of (n+1) using parts 1 mod k. Cf.
for k = 3 and
for k = 4. Example: for k = 2, n = 4 we have the compositions (22; 211, 121; 112; 1111) = 5; but using parts 1 and 3 we have for n = 5: (311, 131, 113, 11111, 5) = 5. - , Jul 05 2012
The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)*F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^n*F(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called &delta-Fibonacci& numbers as defined in comments to
(see also the papers of Witula et al.). - , Jul 24 2012
F(n) is the number of different (n-1)-digit binary numbers such that all substrings of length & 1 have at least one digit equal to 1. Example: for n = 5 there are 8 binary numbers with n - 1 = 4 digits (, , , ), only the F(n) = 5 numbers ,
and 1111 have the desired property. - , Nov 30 2012
For positive n, F(n+1) equals the determinant of the n X n tridiagonal matrix with 1's along the main diagonal, i's along the superdiagonal and along the subdiagonal where i = sqrt(-1). Example: Det([1,i,0,0; i,1,i,0; 0,i,1,i; 0,0,i,1]) = F(4+1) = 5. - , Feb 24 2013
For n&=1, number of compositions of n where there is a drop between every second pair of parts, starting with the f see example. Also, a(n+1) is the number of compositions where there is a drop between every second pair of parts, starting with the s see example. - , May 21 2013
Central terms of triangles in
and , n & 0. - , Jul 28 2013
For n&=4, F(n-1) is the number of simple permutations in the geometric grid class given in . - , Sep 08 2013
a(n) are the pentagon (not pentagonal) numbers because the algebraic degree 2 number rho(5) = 2*cos(Pi/5) = phi (golden section), the length ratio diagonal/side in a pentagon, has minimal polynomial C(5,x) = x^2 - x - 1 (see , n=5), hence rho(5)^n = a(n-1)*1 + a(n)*rho(5), n &= 0, in the power basis of the algebraic number field Q(rho(5)). One needs a(-1) = 1 here. See also the P. Steinbach reference under . - , Oct 01 2013
(a(n)) = 1. - , Oct 10 2013
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n+2k)^2 - F(n)^2 = F(n+k)*( F(n+3k) - F(n-k) ). - , Dec 20 2013
( F(n), F(n+2k) ) satisfies the Diophantine equation: X^2 + Y^2 - L(2k)*X*Y = F(4k)^2*(-1)^n. This generalizes Bouhamida’s comments dated Sep 06 2009 and Sep 08 2009. - , Jan 07 2014
For any prime p there is an infinite periodic subsequence within F(n) divisible by p, that begins at index n = 0 with value 0, and its first nonzero term at n = (i), and period k = (i). Also see . - , Jan 26 2014
Range of row n of the circular Pascal array of order 5. - , May 30 2014 [orig. Kicey-Klimko 2011, and observations by Glen W more general work found in Ault-Kicey 2014]
Nonnegative range of the quintic polynomial 2*y - y^5 + 2*x*y^4 + x^2*y^3 - 2*x^3*y^2 - x^4*y with x, y &= 0, see Jones 1975. - , Jun 01 2014
The expression round(1/(F(k+1)/F(n) + F(k)/F(n+1))), for n & 0, yields a Fibonacci sequence with k-1 leading zeros (with rounding 0.5 to 0). - , Aug 04 2014
Conjecture: For n & 0, F(n) is the number of all admissible residue classes for which specific finite subsequences of the Collatz 3n + 1 function consists of n+2 terms. This has been verified for 0 & n & 51. For details see Links. - , Oct 03 2014
a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - , Oct 02 2014
a(1)=1=a(2), a(3)=2 are the only Fibonacci numbers that are of the form prime-1. - , Jun 07 2015
Any consecutive pair (m, k) of the Fibonacci sequence a(n) illustrates a fair equivalence between m miles and k kilometers. For instance, 8 miles ~ 13 13 miles ~ 21 km. -, Oct 06 2014
(n -& oo) lim (log F(n+1)/log F(n))^n = e. - , Oct 06 2014
a(n+1) counts closed walks on K_2, containing one loop on the other vertex. Equivalently the (1,1)_entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1; 1,1). - , Oct 29 2014
a(n-1) counts closed walks on the graph G(1-l-loop,2-loop). - , Nov 26 2014
From , Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers , with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers,
with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted , the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. ).
BTC(x) = C[Pinv(x)] gives , a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence , so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for , C[P[x,1-t]], and that for , Pinv[Cinv(x),t+1].
In keeping with historical accounts (see the references by P. Singh and S. Kak), the generalized Fibonacci sequence a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, ... can also be described as the Gopala-Hemachandra numbers H(n) = H(n-1) + H(n-2), with F(n) = H(n) for a = b = 1, and Lucas sequence L(n) = H(n) for a = 2, b = 1. - , Jan 11 2015
D. E. Knuth writes: &Before Fibonacci wrote his work, the sequence F_{n} had already been discussed by Indian scholars, who had long been interested in rhythmic patterns that are formed from one-beat and two-beat notes. The number of such rhythms having n beats altogether is F_{n+1}; therefore both Gopāla (before 1135) and Hemachandra (c. 1150) mentioned the numbers 1, 2, 3, 5, 8, 13, 21, ... explicitly.& (TAOCP Vol. 1, 2nd ed.) - , Jan 11 2015
F(n+1) equals the number of binary words of length n avoiding runs of zeroes of odd lengths. - , Jan 28 2015
From , Apr 12 2015: (Start)
We prove Conjecture 1 of Rashid listed in the Formula section.
We use the following notation: F(n)=(n), the Fibonacci numbers, and L(n) = (n), the Lucas numbers. The fundamental Fibonacci-Lucas recursion asserts that G(n) = G(n-1)+ G(n-2), with &L& or &F& replacing &G&.
We need the following prerequisites which we label (A), (B),(C), (D). The prerequisites are formulas in the Koshy book listed in the References section. (A) F(m-1)+F(m+1) = L(m) (Koshy, p. 97, #32), (B) L(2m)+2(-1)^m = L(m)^2 (Koshy p. 97, #41), (C) F(m+k)F(m-k) = (-1)^n F(k)^2 (Koshy, p. 113, #24, Tagiuri's identity), and (D) F(n)^2+F(n+1)^2 = F(2n+1) (Koshy, p. 97, #30).
We must also prove (E), L(n+2) F(n-1) = F(2n+1)+2(-1)^n. To prove (E), first note that by (A), proof of (E) is equivalent to proving that F(n+1)F(n-1) + F(n+3)F(n-1) = F(2n+1)+2(-1)^n. But by (C) with k=1, we have F(n+1)F(n-1) = F(n)^2 +(-1)^n. Applying (C) again with k=2 and m=n+1, we have F(n+3)F(n-1) = F(n+1)+(-1)^n. Adding these two applications of (C) together and using (D) we have, F(n+1)F(n-1) + F(n+3)F(n-1) = F(n)^2 + F(n+1)^2 + 2(-1)^n = F(2n+1)+2(-1)^n, completing the proof of (E).
We now prove Conjecture 1. By (A) and the Fibonacci-Lucas recursion, we have F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4) = [F(2n+1)+F(2n+3)] + [F(2n+2)+F(2n+4)] = L(2n+2)+L(2n+3)=L(2n+4). But then by (B), with m=2n+4, we have sqrt(L(2n+4)+2(-1)^n)) = L(n+2). Finally by (E), we have L(n+2) F(n-1)= F(2n+1)+2*(-1)^n. Dividing both sides by F(n-1), we have (F(2n+1)+2*(-1)^n)/F(n-1) = L(n+2) = sqrt(F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4)+2(-1)^n), as required.
In Fibonacci's Liber Abaci the rabbit problem appears in the translation of L. E. Sigler on pp. 404-405, and a remark [27] on p. 637. - , Apr 17 2015
a(n) counts partially ordered partitions of (n-1) into parts 1,2,3 where only the order of adjacent 1's and 2's are unimportant. (See example.) - , Jul 27 2015
F(n) divides F(nk). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - , Aug 24 2015
F(n) is the number of UDU-equivalence classes of ballot paths of length n. Two ballot paths of length n with steps U = (1,1), D = (1,-1) are UDU-equivalent whenever the positions of UDU are the same in both paths. - , Aug 25 2015
Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1 is the basis for a geometrical paradox (or dissection fallacy) in . - , Oct 23 2015
For n &= 4, F(n) is the number of up-down words on alphabet {1,2,3} of length n-2. - , Nov 23 2015
F(n+2) is the number of terms in p(n), where p(n)/q(n) is the n-th convergent of the formal infinite continued fraction [a(0),a(1),...]; e.g., p(3) = a(0)a(1)a(2)a(3) + a(0)a(1) + a(0)a(3) + a(2)a(3) + 1 has F(5) terms. Also, F(n+1) is the number of terms in q(n). - , Dec 23 2015
F(n+1) (for n&=1) is the permanent of an n X n matrix M with M(i,j)=1 if |i-j|&=1 and 0 otherwise. - , Jan 08 2016
A trapezoid has three sides of lengths in order F(n), F(n+2), F(n). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*F(n+1). For a trapezoid with lengths of sides in order F(n+2), F(n), F(n+2), the fourth side will be F(n+3). - , Mar 17 2016
(1) Join two triangles with lengths of sides L(n), F(n+3), L(n+2) and F(n+2), L(n+1), L(n+2) (where L(n)=(n)) along the common side of length L(n+2) to create an irregular quadrilateral. Its area is approximately (5*F(2*n-1) - (F(2*n-7) - F(2*n-13))/5. (2) Join two triangles with lengths of sides L(n), F(n+2), F(n+3) and L(n+1), F(n+1, F(n+3) along the common side F(n+3) to form an irregular quadrilateral. Its area is approximately 4*F(2*n-1) - 2*(F(2*n-7) + F(2*n-18)). - , Apr 06 2016
From , Jun 13 2016: (Start)
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2, x}, g(3) = {3, 2x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If a positive integer N is not a square and r = sqrt(N), then the number of (not necessarily distinct) integers in g(n) is (n), for n & = 1. See . (End)
Consider the partitions of n, with all summands initially listed in nonincreasing order. Freeze all the 1's in place and then allow all the other summands to change their order, without displacing any of the 1's. The resulting number of arrangements is a(n+1). - , Jun 14 2016
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, signature (1,1).
G.f.: x / (1 - x - x^2).
G.f.: Sum_{n&=0} x^n * Product_{k=1..n} (k + x)/(1 + k*x). - , Oct 26 2013
F(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5)).
Alternatively, F(n) = ((1/2+sqrt(5)/2)^n-(1/2-sqrt(5)/2)^n)/sqrt(5).
F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n).
F(n) = round(phi^n/sqrt(5)).
F(n+1) = Sum_{j=0..[n/2]} binomial(n-j, j).
This is a d that is, if n divides m, then a(n) divides a(m). - , Apr 07 2012
E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2). - , Nov 30 2001
[0 1; 1 1]^n [0 1] = [F(n); F(n+1)]
x | F(n) ==& x | F(kn).
A sufficient condition for F(m) to be divisible by a prime p is (p - 1) divides m, if p == 1 or 4 (mod 5); (p + 1) divides m, if p == 2 or 3 (mod 5); or 5 divides m, if p = 5. (This is essentially Theorem 180 in Hardy and Wright.) - Fred W. Helenius (fredh(AT)), Jun 29 2001
a(n)=F(n) has the property: F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1). - , Nov 13 2003
, Feb 21 2004, makes 4 conjectures and gives 3 theorems:
Conjecture 1: for n&=2 sqrt{F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4)+2*(-1)^n}={F(2n+1)+2*(-1)^n}/F(n-1). [For a proof see Comments section.]
Conjecture 2: for n&=0, {F(n+2)* F(n+3)}-{F(n+1)* F(n+4)}+ (-1)^n = 0.
Conjecture 3: for n&=0, F(2n+1)^3 - F(2n+1)*[(2*A^2)-1] - [A + A^3]=0, where A = {F(2n+1)+sqrt{5*F(2n+1)^2 +4}}/2.
Conjecture 4: for x&=5, if x is a Fibonacci number &= 5 then g*x*[{x+sqrt{5*(x^2) +- 4}}/2]*[2x+{{x+sqrt{5*(x^2) +- 4}}/2}]*[2x+{{3x+3*sqrt {5*(x^2) +- 4}}/2}]^2+[2x+{{x+sqrt{5*(x^2) +- 4}}/2}] +- x*[2x+{{3x+3*sqrt{5*(x^2) +- 4}}/2}]^2 -x*[2x+{{x+sqrt{5*(x^2) +- 4}}/2}]*[x+{{x+sqrt{5*(x^2) +- 4}}/2}]* [2x+{{3x+3*sqrt{5*(x^2) +- 4}}/2}]^2= 0, where g = {1 + sqrt(5)/2}.
Theorem 1: for n&=0, {F(n+3)^ 2 - F(n+1)^ 2}/F(n+2)={F(n+3)+ F(n+1)}.
Theorem 2: for n&=0, F(n+10) = 11*F(n+5) + F(n).
Theorem 3: for n&=6, F(n) = 4*F(n-3) + F(n-6).
Conjecture 2 of Rashid is actually a special case of the general law F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1) (take n &- n+1 and m &- -(n+4) in this law). - Harmel Nestra (harmel.nestra(AT)ut.ee), Apr 22 2005
Conjecture 2 of Rashid Kurmang simplified: F(n)*F(n+3) = F(n+1)*F(n+2)-(-1)^n. Follows from d'Ocagne's identity: m=n+2. - , May 06 2012
Conjecture: for all c such that 2-Phi &= c & 2*(2-Phi) we have F(n) = floor(Phi*a(n-1)+c) for n & 2. - , Jul 21 2004
|2*Fib(n) - 9*Fib(n+1)| = 4*(n) + (n+1). - , Aug 13 2004
For x & Phi, Sum_{n=0..inf} F(n)/x^n = x/(x^2 - x - 1) - , Oct 27 2004
F(n+1) = exponent of the n-th term in the series f(x, 1) determined by the equation f(x, y) = xy + f(xy, x). - , Dec 19 2004
a(n-1) = Sum_{k=0..n} (-1)^k*binomial(n-ceil(k/2), floor(k/2)). - , May 05 2005
F(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)(1+(-1)^(n-k))/2. - , Aug 28 2005
Fibonacci(n) = Product(1 + 4[cos(j*Pi/n)]^2, j=1..ceil(n/2)-1). [Bicknell and Hoggatt, pp. 47-48.] - , Oct 15 2006
F(n) = 2^-(n-1)*Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)*5^k. - , Feb 07 2006
a(n) = (b(n+1)+b(n-1))/n where {b(n)} is the sequence . - , Nov 22 2006
F(n*m) = Sum_{k = 0..m} binomial(m,k)*F(n-1)^k*F(n)^(m-k)*F(m-k). The generating function of F(n*m) (n fixed, m = 0,1,2...) is G(x) = F(n)*x / ((1-F (n-1)*x)^2-F(n)*x*(1-F(n-1)*x)-( F(n)*x)^2). E.g., F(15) = 610 = F(5*3) = binomial(3,0)* F(4)^0*F(5)^3*F(3) + binomial(3,1)* F(4)^1*F(5)^2*F(2) + binomial(3,2)* F(4)^2*F(5)^1*F(1) + binomial(3,3)* F(4)^3*F(5)^0*F(0) = 1*1*125*2 + 3*3*25*1 + 3*9*5*1 + 1*27*1*0 = 250 + 225 + 135 + 0 = 610. - , Feb 12 2007
From , Mar 19 2007: (Start)
&&Let L(n) = (n) = Lucas numbers. Then:
&&For a&=b and odd b, F(a+b)+F(a-b)=L(a)*F(b).
&&For a&=b and even b, F(a+b)+F(a-b)=F(a)*L(b).
&&For a&=b and odd b, F(a+b)-F(a-b)=F(a)*L(b).
&&For a&=b and even b, F(a+b)-F(a-b)=L(a)*F(b).
&&F(n+m)+(-1)^m*F(n-m)=F(n)*L(m);
&&F(n+m)-(-1)^m*F(n-m)=L(n)*F(m);
&&F(n+m+k)+(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k)) =F(n)*L(m)*L(k);
&&F(n+m+k)-(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k)) =L(n)*L(m)*F(k);
&&F(n+m+k)+(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k)) =L(n)*F(m)*L(k);
&&F(n+m+k)-(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k)) =5*F(n)*F(m)*F(k). (End)
A corollary to Kristof 2007 is 2*F(a+b)=F(a)*L(b)+L(a)*F(b). - , Apr 24 2014
For n&m, the sum of the 2m consecutive Fibonacci numbers F(n-m-1) thru F(n+m-2) is F(n)*L(m) if m is odd, and L(n)*F(m) if m is even (see the McRae link). - , Apr 24 2014.
Fib(n) = b(n)+(p-1)*Sum_{1&k&n} floor(b(k)/p)*Fib(n-k+1) where b(k) is the digital sum analog of the Fibonacci recurrence, defined by b(k)=ds_p(b(k-1))+ds_p(b(k-2)), b(0)=0, b(1)=1, ds_p=digital sum base p. Example for base p=10: Fib(n)=(n)+9*Sum_{1&k&n} ((k))*Fib(n-k+1). - , Jul 01 2007
Fib(n) = b(n)+p*Sum_{1&k&n} floor(b(k)/p)*Fib(n-k+1) where b(k) is the digital product analog of the Fibonacci recurrence, defined by b(k)=dp_p(b(k-1))+dp_p(b(k-2)), b(0)=0, b(1)=1, dp_p=digital product base p. Example for base p=10: Fib(n)=(n)+10*Sum_{1&k&n} ((k))*Fib(n-k+1). - , Jul 01 2007
a(n) = denominator of continued fraction [1,1,1,...] (with n ones); e.g., 2/3 = continued fraction [1,1,1]; where barover[1] = [1,1,1...] = 0.6180339.... - , Nov 29 2007
F(n + 3) = 2F(n + 2) - F(n), F(n + 4) = 3F(n + 2) - F(n), F(n + 8) = 7F(n + 4) - F(n), F(n + 12) = 18F(n + 6) - F(n). - , Feb 01 2008
1 = 1/(1*2) + 1/(1*3) + 1/(2*5) + 1/(3*8) + 1/(5*13) + ... = 1/2 + 1/3 + 1/10 + 1/24 + 1/65 + 1/168 + ...; where
= (0, 2, 3, 10, 24, 65, 168,...). - , Mar 16 2008
a(2^n) = prod{i=0}^{n-2}B(i) where B(i) is . Example 3*7*47 = Fib(16). - , Apr 23 2008
F(n) = (1/(n-1)!) * (n^(n-1) - (C(n-2,0) + 4*C(n-2,1) + 3*C(n-2,2))*n^(n-2) + (10*C(n-3,0) + 49*C(n-3,1) + 95*C(n-3,2) + 83*C(n-3,3) + 27*C(n-3,4))*n^(n-3) - (90*C(n-4,0) + 740*C(n-4,1) + 2415*C(n-4,2) + 4110*C(n-4,3) + 3890*C(n-4,4) + 1950*C(n-4,5) + 405*C(n-4,6))*n^(n-4) + ... ). - , Nov 24 2004
a(n+1) = Sum_{k, 0&=k&=n} (n,k)*(-1)^(n-k). -, Oct 26 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}... Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n), where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i + l_(i+1) &= 2 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - , Feb 25 2009
a(n+1) = 2^n sqrt(Product_{k=1..n} cos(k Pi/(n+1))^2+1/4)) (Kasteleyn's formula specialized). - Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009
a(n+1) = Sum_{k=floor[n/2] mod 5} C(n,k) - Sum_{k=floor[(n+5)/2] mod 5} C(n,k) = (n) - (n) = |(n)-(n-1)|. - , Feb 10 2010
If p[i]=modp(i,2) and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i&=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n&=1, a(n)=det A. - , May 02 2010
Limit(F(k+n)/F(k), k = infinity) = (L(n) + F(n)*sqrt(5))/2 with the Lucas numbers L(n)= (n). - , May 27 2010
For n&=1, F(n)=round(log_2(2^{phi*F(n-1)} + 2^{phi*F(n-2)})), where phi is the golden ratio. - , Jun 24 2010, Jun 27 2010
For n&=1, a(n+1)=ceil(phi*a(n)), if n is even and a(n+1)=floor(phi*a(n)), if n is odd (phi = golden ratio). - , Jul 01 2010
a(n) = 2*a(n-2) + a(n-3), n&2. - , Sep 08 2010
a(2^n) = Prod_{i=0..n-1} (2^i). - , Nov 28 2010
a(n)^2 - a(n-1)^2 = a(n+1)*a(n-2), see .
a(n) = sqrt((-1)^k*(a(n+k)^2 - a(k)*a(2n+k))), for any k. - , Dec 03 2010
F(2*n) = F(n+2)^2 - F(n+1)^2 - 2*F(n)^2. - , Jun 04 2011
(-1)^(n+1) = F(n)^2 + F(n)*F(1+n) - F(1+n)^2.
&&F(n) = -F(n+2)(-2 + (F(n+1))^4 + 2*(F(n+1)^3*F(n+2)) - (F(n+1)*F(n+2))^2 2*F(n+1)(F(n+2))^3 + (F(n+2))^4)- F(n+1). - , Nov 17 2011
F(n) = 1 + Sum_{x=1..n-2} F(x). - , Feb 05 2012
F(n) = 4*F(n-2) - 2*F(n-3) - F(n-6). - , Apr 01 2012
F(n) = round(phi^(n+1)/(phi+2)). - , Apr 20 2012
From , Jun 03 2012: (Start)
G.f. A(x) = x/(1-x-x^2) = G(0)/sqrt(5) where G(k)= 1 -((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*c^k/G(k+1))) and a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
Let E(x) be the e.g.f., i.e.,
E(x) = 1*x + 1/2*x^2 + 1/3*x^3 + 1/8*x^4 + 1/24*x^5 + 1/90*x^6 + 13/5040*x^7 + ...; then
E(x) = G(0)/sqrt(5); G(k)= 1 -((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*(k+1)*c^k/G(k+1))), where a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
From , Nov 30 2012: (Start)
Fib(n) = 1 + Sum_{j_1=1..n-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} 1 + ... + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} ... Sum_{j_k=1..j_(k-1)-2} 1, where k = floor((n-1)/2).
Example: Fib(6) = 1 + Sum_{j=1..4} 1 + Sum_{j=1..4} Sum_{k=1..(j-2)} 1 + 0 = 1 + (1 + 1 + 1 + 1) + (1 + (1 + 1)) = 8.
Fib(n) = Sum_{j=0..k} S(j+1,n-2j), where k = floor((n-1)/2) and the S(j,n) are the n-th j-simplex sums: S(1,n) = 1 is the 1-simplex sum, S(2,n) = Sum_{k=1..n} S(1,k) = 1+1+...+1 = n is the 2-simplex sum, S(3,n) = Sum_{k=1..n} S(2,k) = 1+2+3+...+n is the 3-simplex sum (= triangular numbers = ), S(4,n) = Sum_{k=1..n} S(3,k) = 1+3+6+...+n(n+1)/2 is the 4-simplex sum (= tetrahedral numbers = ) and so on.
Since S(j,n) = binomial(n-2+j,j-1), the formula above equals the well-known binomial formula, essentially. (End)
G.f. A(x) = x / (1 - x / (1 - x / (1 + x))). - , Jan 04 2013
Sum_{n&=1} (-1)^(n-1)/(a(n)*a(n+1)) = 1/phi (phi=golden ratio). - , Feb 22 2013
From , Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (a(n) + sqrt(5*(a(n))^2 + 4*(-1)^n))/2;
(2) Sum_{k=1...n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(3) a(n)/a(n+1) = 1/phi + r(n), where |r(n)| & 1/(a(n+1)*a(n+2)). (End)
F(n+1) = F(n)/2 + sqrt((-1)^n + 5*F(n)^2/4), n&=0. F(n+1) = U_n(i/2)/i^n, (U:= Chebyshef 2nd kind). - , Mar 04 2013
G.f.: -Q(0) where Q(k) = 1 - (1+x)/(1 - x/(x - 1/Q(k+1) )); (continued fraction). - , Mar 06 2013
G.f.: x-1-1/x + 1/x/Q(0), where Q(k) = 1 - (k+1)*x/(1 - x/(x - (k+1)/Q(k+1))); (continued fraction). - , Apr 23 2013
G.f.: x*G(0), where G(k)= 1 + x*(1+x)/(1 - x*(1+x)/(x*(1+x) + 1/G(k+1) )); (continued fraction). - , Jul 08 2013
G.f.: x^2 - 1 + 2*x^2/(W(0)-2), where W(k) = 1 + 1/(1 - x*(k + x)/( x*(k+1 + x) + 1/W(k+1) )); (continued fraction). - , Aug 28 2013
G.f.: Q(0) -1, where Q(k) = 1 + x^2 + (k+2)*x -x*(k+1 + x)/Q(k+1); (continued fraction). - , Oct 06 2013
Let b(n) = b(n-1) + b(n-2), with b(0) = 0, b(1) = phi. Then, for n&=2, F(n)= floor(b(n-1)) if n is even, F(n) = ceil(b(n-1)), if n is odd, with convergence. - , Jan 19 2014
a(n) = Sum_{t1*g(1)+t2*g(2)+...+tn*g(n)=n} multinomial(t1+t2 +...+tn,t1,t2,...,tn), where g(k)=2*k-1. - , Feb 27 2014
F(n) = round(sqrt(F(n-1)^2 + F(n)^2 + F(n+1)^2)/2), for n & 0. This rule appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - , Jul 27 2014
F(n) = round(2/(1/F(n) + 1/F(n+1) + 1/F(n+2)), for n & 0. This rule also appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - , Aug 03 2014
F(n) = round(1/(Sum_{j&=n+2} 1/F(j))). - , Aug 14 2014
a(n) = hypergeometric([-n/2+1/2, -n/2+1], [-n+1], -4) for n&=2. - , Sep 19 2014
F(n) = (L(n+1)^2 - L(n-1)^2)/(5*L(n)), where L(n) is (n), with a similar inverse relationship. - , Nov 17 2014
Consider the graph G[1-1-loop,2-loop] in comment above. Construct the power matrix array T(n,j)=[A^*j]*[S^*(j-1)] where A=(1,1,0,...) and S=(0,1,0,...)(). [* is convolution operation] Define S^*0=I with I=(1,0,...). Then T(n,j) counts n-walks containing (j) loops and a(n-1) = Sum_{j=1...n} T(n,j). - , Nov 21 2014
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n) = F(k)*F(n-k+3) - F(k-1)*F(n-k+2) - F(k-2)*F(n-k) + (-1)^k*F(n-2k+2). - , Dec 04 2014
F(n+k)^2 - L(k)*F(n)*F(n+k) + (-1)^k*F(n)^2 = (-1)^n*F(k)^2, if L(k) = (k). - , Jul 20 2015
F(2*n) = F(n+1)^2 - F(n-1)^2, similar to Koshy (D) and Forberg 2011, but different. - , Aug 12 2015
F(n+1) = ceiling( (1/phi)*Sum_{k=0..n} F(k) ). - , Sep 10 2015
a(n) = (L(n-3) + L(n+3))/10 where L(n)=(n). - , Nov 25 2015
From , Mar 27 2016 (Start):
F(n) = (F(2n+k+1) - F(n+1)*F(n+k+1))/F(n+k), k&=0.
Thus when k=0: F(n) = sqrt(F(2n+1) - F(n+1)^2).
F(n) = cbrt(F(3n) - F(n+1)^3 + F(n-1)^3).
F(n+2k) = binomial transform of any subsequence starting with F(n). Example F(6)=8: 1*8 = F(6)=8; 1*8 + 1*13 = F(8)=21; 1*8 + 2*13 + 1*21 = F(10)=55; 1*8 + 3*13 + 3*21 + 1*34 = F(12)=144, etc. This formula applies to Fibonacci-type sequences with any two seed values for a(0) and a(1) (e.g., Lucas sequence : a(0)=2, a(1)=1).
For x = 0,1,2,3,4, x=1/(x+1) = 1, 1/2, 2/3, 3/5, 5/8. These fractions have numerators 1,1,2,3,5, which are the 2nd to 6th entries in the sequence. - , Sep 15 2008
From , May 21 2013: (Start)
There are a(7)=13 compositions of 7 where there is a drop between every second pair of parts, starting with the first and second part:
01:&&[ 2 1 2 1 1 ]
02:&&[ 2 1 3 1 ]
03:&&[ 2 1 4 ]
04:&&[ 3 1 2 1 ]
05:&&[ 3 1 3 ]
06:&&[ 3 2 2 ]
07:&&[ 4 1 2 ]
08:&&[ 4 2 1 ]
09:&&[ 4 3 ]
10:&&[ 5 1 1 ]
11:&&[ 5 2 ]
12:&&[ 6 1 ]
13:&&[ 7 ]
There are abs(a(6+1))=13 compositions of 6 where there is no rise between every second pair of parts, starting with the second and third part:
01:&&[ 1 2 1 2 ]
02:&&[ 1 3 1 1 ]
03:&&[ 1 3 2 ]
04:&&[ 1 4 1 ]
05:&&[ 1 5 ]
06:&&[ 2 2 1 1 ]
07:&&[ 2 3 1 ]
08:&&[ 2 4 ]
09:&&[ 3 2 1 ]
10:&&[ 3 3 ]
11:&&[ 4 2 ]
12:&&[ 5 1 ]
13:&&[ 6 ]
Partially ordered partitions of (n-1) into parts 1,2,3 where only the order of the adjacent 1's and 2's are unimportant. E.g., a(8)=21. These are (331),(313),(133),(322),(232),(223),(3211),(2311),(1321),(2131),(1132),(2113),(31111),(13111),(11311),(11131),(11113),(2221),(22111),(211111),(1111111). - , Jul 25 2015
Consider the partitions of 7 with summands initially listed in nonincreasing order. Keep the 1's frozen in position,(indicated by &[]&) and then allow the other summands to otherwise vary their order: 7; 6,[1]; 5,2; 2,5; 4,3; 3,4; 5,[1,1], 4,2,[1]; 2,4,[1]; 3,3,[1]; 3,3,2; 3,2,3; 2,3,3; 4,[1,1,1]; 3,2,[1,1]; 2,3,[1,1]; 2,2,2,[1]; 3,[1,1,1,1]; 2,2,[1,1,1]; 2,[1,1,1,1,1]; [1,1,1,1,1,1,1]. There are 21 = a(7+1) arrangements in all. - , Jun 14 2016
:= proc(n) combinat[fibonacci](n);
ZL:=[S, {a = Atom, b = Atom, S = Prod(X, Sequence(Prod(X, b))), X = Sequence(b, card &= 1)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=0..38); # , Apr 04 2008
spec := [B, {B=Sequence(Set(Z, card&1))}, unlabeled ]: seq(combstruct[count](spec, size=n), n=1..39); # , Apr 04 2008
# The following Maple command isFib(n) yields true or false depending on whether n is a Fibonacci number or not.
with(combinat): isFib := proc(n) local a: a := proc(n) local j: for j while fibonacci(j) &= n do fibonacci(j) end do: fibonacci(j-1) end proc: evalb(a(n) = n) end proc: # , Nov 11 2014
MATHEMATICA
Table[ Fibonacci[ k ], {k, 0, 50} ] (* to generate 0 as well as all other nonzero Fibonacci numbers, the lower limit is changed from 1 to 0; , Jul 11 2015 *)
Table[ 2^n Sqrt@Product[( Cos[Pi k/(n + 1)]^2 + 1/4), {k, n}]//FullSimplify, {n, 15}]; (* Kasteleyn's formula specialized, Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009 *)
Table[Fibonacci[n]^5 - Fibonacci[1 + n] + 3 Fibonacci[n]^4 Fibonacci[1 + n] + Fibonacci[n]^3 Fibonacci[1 + n]^2 - 3 Fibonacci[n]^2 Fibonacci[1 + n]^3 - Fibonacci[n] Fibonacci[1 + n]^4 + Fibonacci[1 + n]^5, {n, 1, 10}] (* , Nov 17 2011 *)
LinearRecurrence[{1, 1}, {0, 1}, 40] (* , Aug 03 2014 *)
(Axiom) [fibonacci(n) for n in 0..50]
(MAGMA) [Fibonacci(n): n in [0..38]];
(Maxima) makelist(fib(n), n, 0, 100); /* , Oct 21 2012 */
(PARI) a(n) = fibonacci(n)
(PARI) a(n) = imag(quadgen(5)^n)
(PARI) a(n)=my(phi=quadgen(5)); (phi^n-(-1/phi)^n)/(2*phi-1) \\ , Jun 17 2012
(PARI) a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, k+x +x*O(x^n))/prod(k=1, m, 1+k*x +x*O(x^n))), n) \\ , Oct 26 2013
(Python) # , Jan 05 2007 (Change leading dots to blanks.)
def fib():
... &&& Generates the Fibonacci numbers, starting with 0 &&&
... x, y = 0, 1
... while 1:
....... yield x
....... x, y = y, x+y
a = [f.next() for i in range(100)]
... &&& Returns Fibonacci number with index n, offset 0, 4 &&&
... return a[n]
................
def _list(N):
... &&& Returns a list of the first n Fibonacci numbers &&&
... return a[:N]
(Sage) ## Demonstration program from Jaap Spies:
a = sloane.; ## choose sequence
print a ## This returns the name of the sequence.
print a(38) ## This returns the 38th number of the sequence.
print a.list(39) ## This returns a list of the first 39 numbers.
-- Based on code from http://www.haskell.org/haskellwiki/The_Fibonacci_sequence
-- which also has other versions.
fib :: Int -& Integer
fib n = fibs !! n
.... fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
{- Example of use: map fib [0..38] , Sep 29 2009 -}
(Sage) [i for i in fibonacci_sequence(0, 40)] # , Jun 26 2014
(MAGMA) [0, 1] cat [n: n in [1..] | IsSquare(5*n^2-4) or IsSquare(5*n^2+4)]; // , Nov 19 2014
(signed Fibonacci numbers),
(complement), , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , .
First row of arrays , . Second row of arrays , , and
(k-generalized Fibonacci numbers).
a(n) = (4, n). a(n) = (0, j, n).
a(n) = (0, n+1) = (0, n+1) = (1, n-1) = (0, n) = (1, n-1), for n & 0.
Fibonacci-Pascal triangles: , , , , , , , , , .
Boustrophedon transforms: , .
Powers: , , .
Sequence in context:
Adjacent sequences:&&
core,nonn,nice,easy,,changed
, Apr 30 1991
Maintained by
Last modified August
4 18:06 EDT 2016.
Contains 275300 sequences.}
请问：y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))printf()函数中’\n’；’\t’；’\a
B我们讨论的范围是x&0\AD=AB BD=AB BC/2
提问者采纳
x-1BE=BC CE=BC CA/0所以f(x)=1
b f(x)=x-1
g(x)=x^2/0所以|x-1| |x-2|>mx2-(1-m)x 1>
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出门在外也不愁您的位置： >>
好吧，我承认我标题党了，不过既然你来了，就认真看下去吧，保证你有收获。
我们平时经常会有一些数据运算的操作，需要调用sqrt，exp，abs等函数，那么时候你有没有想过：这个些函数系统是如何实现的？就拿最常用的sqrt函数来说吧，系统怎么来实现这个经常调用的函数呢？
虽然有可能你平时没有想过这个问题，不过正所谓是&临阵磨枪，不快也光&，你&眉头一皱，计上心来&，这个不是太简单了嘛，用二分的方法，在一个区间中，每次拿中间数的平方来试验，如果大了，就再试左区间的中间数；如果小了，就再拿右区间的中间数来试。比如求sqrt(16)的结果，你先试(0+16)/2=8，8*8=64，64比16大，然后就向左移，试(0+8)/2=4，4*4=16刚好，你得到了正确的结果sqrt(16)=4。然后你三下五除二就把程序写出来了：
float SqrtByBisection(float n) //用二分法
if(n&0) //小于0的按照你需要的处理
float mid,
float low,
low=0, up=n;
mid=(low+up)/2;
if(mid*mid&n)
mid=(up+low)/2;
}while(abs(mid-last) & eps);//精度控制
然后看看和系统函数性能和精度的差别（其中时间单位不是秒也不是毫秒，而是CPU Tick，不管单位是什么，统一了就有可比性）&
从图中可以看出，二分法和系统的方法结果上完全相同，但是性能上整整差了几百倍。为什么会有这么大的区别呢？难道系统有什么更好的办法？难道。。。。哦，对了，回忆下我们曾经的高数课，曾经老师教过我们&牛顿迭代法快速寻找平方根&，或者这种方法可以帮助我们，具体步骤如下：
求出根号a的近似值：首先随便猜一个近似值x，然后不断令x等于x和a/x的平均数，迭代个六七次后x的值就已经相当精确了。 例如，我想求根号2等于多少。假如我猜测的结果为4，虽然错的离谱，但你可以看到使用牛顿迭代法后这个值很快就趋近于根号2了： (4 + 2/4) / 2 = 2.25 (2.25 + 2/2.25) / 2 = 1.56944.. (1.56944.. + 2/1.56944..) / 2 = 1.42189.. (1.42189.. + 2/1.42189..) / 2 = 1.41423.. ....
这种算法的原理很简单，我们仅仅是不断用(x,f(x))的切线来逼近方程x^2-a=0的根。根号a实际上就是x^2-a=0的一个正实根，这个函数的导数是2x。也就是说，函数上任一点(x,f(x))处的切线斜率是2x。那么，x-f(x)/(2x)就是一个比x更接近的近似值。代入 f(x)=x^2-a得到x-(x^2-a)/(2x)，也就是(x+a/x)/2。
相关的代码如下：
float SqrtByNewton(float x)
float val = //最终
float //保存上一个计算的值
val =(val + x/val) / 2;
}while(abs(val-last) & eps);
然后我们再来看下性能测试：
哇塞，性能提高了很多，可是和系统函数相比，还是有这么大差距，这是为什么呀？想啊想啊，想了很久仍然百思不得其解。突然有一天，我在网上看到一个神奇的方法，于是就有了今天的这篇文章，废话不多说，看代码先：
float InvSqrt(float x)
float xhalf = 0.5f*x;
int i = *(int*)&x; // get bits for floating VALUE
i = 0x5f375a86- (i&&1); // gives initial guess y0
x = *(float*)&i; // convert bits BACK to float
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
return 1/x;
然后我们最后一次来看下性能测试：
这次真的是质变了，结果竟然比系统的还要好。。。哥真的是震惊了！！！哥吐血了！！！一个函数引发了血案！！！血案，血案。。。
到现在你是不是还不明白那个&鬼函数&，到底为什么速度那么快吗？不急，先看看下面的故事吧：
Quake-III Arena (雷神之锤3)是90年代的经典游戏之一。该系列的游戏不但画面和内容不错，而且即使计算机配置低，也能极其流畅地运行。这要归功于它3D引擎的开发者约翰&卡马克（John Carmack）。事实上早在90年代初DOS时代，只要能在PC上搞个小动画都能让人惊叹一番的时候，John Carmack就推出了石破天惊的Castle Wolfstein，然后再接再励，doom, doomII, Quake...每次都把3-D技术推到极致。他的3D引擎代码资极度高效，几乎是在压榨PC机的每条运算指令。当初MS的Direct3D也得听取他的意见，修改了不少API。
最近，QUAKE的开发商ID SOFTWARE 遵守GPL协议，公开了QUAKE-III的原代码，让世人有幸目睹Carmack传奇的3D引擎的原码。这是QUAKE-III原代码的下载地址：
(下面是官方的下载网址，搜索 &quake3-1.32b-source.zip& 可以找到一大堆中文网页的。
我们知道，越底层的函数，调用越频繁。3D引擎归根到底还是数学运算。那么找到最底层的数学运算函数（在game/code/q_math.c）， 必然是精心编写的。里面有很多有趣的函数，很多都令人惊奇，估计我们几年时间都学不完。在game/code/q_math.c里发现了这样一段代码。它的作用是将一个数开平方并取倒，经测试这段代码比(float)(1.0/sqrt(x))快4倍：
float Q_rsqrt( float number )
const float threehalfs = 1.5F;
x2 = number * 0.5F;
= * ( long * ) &y;
// evil floating point bit level hacking
= 0x5f3759df - ( i && 1 ); // what the fuck?
= * ( float * ) &i;
= y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
= y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
#ifndef Q3_VM
#ifdef __linux__
assert( !isnan(y) ); // bk010122 - FPE?
函数返回1/sqrt(x)，这个函数在图像处理中比sqrt(x)更有用。&
注意到这个函数只用了一次叠代！（其实就是根本没用叠代，直接运算）。编译，实验，这个函数不仅工作的很好，而且比标准的sqrt()函数快4倍！要知道，编译器自带的函数，可是经过严格仔细的汇编优化的啊！
这个简洁的函数，最核心，也是最让人费解的，就是标注了&what the fuck?&的一句：
i = 0x5f3759df - ( i && 1 );
= y * ( threehalfs - ( x2 * y * y ) );
两句话就完成了开方运算！而且注意到，核心那句是定点移位运算，速度极快！特别在很多没有乘法指令的RISC结构CPU上，这样做是极其高效的。
算法的原理其实不复杂，就是牛顿迭代法，用x-f(x)/f'(x)来不断的逼近f(x)=a的根。
没错，一般的求平方根都是这么循环迭代算的，但是卡马克(quake3作者)真正牛B的地方是他选择了一个神秘的常数 0x5f3759df 来计算那个猜测值，就是我们加注释的那一行，那一行算出的值非常接近1/sqrt(n)，这样我们只需要2次牛顿迭代就可以达到我们所需要的精度。好吧如果这个还不算NB，接着看:
普渡大学的数学家Chris Lomont看了以后觉得有趣，决定要研究一下卡马克弄出来的这个猜测值有什么奥秘。Lomont也是个牛人，在精心研究之后从理论上也推导出一个最佳猜测值，和卡马克的数字非常接近，0x5f37642f。卡马克真牛，他是外星人吗？
传奇并没有在这里结束。Lomont计算出结果以后非常满意，于是拿自己计算出的起始值和卡马克的神秘数字做比赛，看看谁的数字能够更快更精确的求得平方根。结果是卡马克赢了... 谁也不知道卡马克是怎么找到这个数字的。
最后Lomont怒了，采用暴力方法一个数字一个数字试过来，终于找到一个比卡马克数字要好上那么一丁点的数字，虽然实际上这两个数字所产生的结果非常近似，这个暴力得出的数字是0x5f375a86。
Lomont为此写下一篇论文，"Fast Inverse Square Root"。 论文下载地址：
参考：&IEEE Standard 754 for Binary Floating-Point Arithmetic&&FAST INVERSE SQUARE ROOT&
最后，给出最精简的1/sqrt()函数：
float InvSqrt(float x)
float xhalf = 0.5f*x;
int i = *(int*)&x; // get bits for floating VALUE
i = 0x5f375a86- (i&&1); // gives initial guess y0
x = *(float*)&i; // convert bits BACK to float
x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy
大家可以尝试在PC机、51、AVR、430、ARM、上面编译并实验，惊讶一下它的工作效率。
前两天有一则新闻，大意是说 Ryszard Sommefeldt 很久以前看到这么样的一段 code (可能出自 Quake III 的 source code)：
float InvSqrt (float x)
float xhalf = 0.5f*x;
int i = *(int*)&x;
i = 0x5f3759df - (i&&1);
x = *(float*)&i;
x = x*(1.5f - xhalf*x*x);
他一看之下惊为天人，想要拜见这位前辈高人，但是一路追寻下去却一直找不到人；同时间也有其他人在找，虽然也没找到出处，但是 Chris Lomont 写了一篇论文 (in PDF) 解析这段 code 的算法 (用的是 Newton&s Method，牛顿法；比较重要的是后半段讲到怎么找出神奇的 0x5f3759df 的)。&
PS. 这个 function 之所以重要，是因为求&开根号倒数&这个动作在 3D 运算 (向量运算的部份) 里面常常会用到，如果你用最原始的 sqrt() 然后再倒数的话，速度比上面的这个版本大概慢了四倍吧& XD PS2. 在他们追寻的过程中，有人提到一份叫做 MIT HACKMEM 的文件，这是 1970 年代的 MIT 强者们做的一些笔记 (hack memo)，大部份是 algorithm，有些 code 是 PDP-10 asm 写的，另外有少数是 C code (有人整理了一份列表)
好了，故事就到这里结束了，希望大家能有有收获:)
编程基础热门文章
编程基础最新文章请问一下y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))AB=AC=3COS=1/9_百度知道
请问一下y=a1.*exp(-sqrt((a2-lab)^2/a5 (a3-cap)^2/a6 (a4-xin)^2/a7))AB=AC=3COS=1/9
oga[(x2^2-ax2)/0f(x)=x^5/5-ax^3/(x1^2-ax1)]>
提问者采纳
2<,0)-1/2;x1<AB BC CA)/x2&2=0f(x)=loga(x^2-ax)在(-1/0仿照线相交于点O仿照AB BC CA)/
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出门在外也不愁(Greetings from !)
Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
(Formerly M)
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, , , 1, 2, 7, , 49, 05, 088169
Also sometimes called Lamé's sequence.
F(n+2) = number of binary sequences of length n that have no consecutive 0's.
F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.
F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.
F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - , Jun 18 2001
F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]
Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y &= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z&0 then z=F(n + 1).
For Fibonacci search see Knuth, Vol. 3; Horowitz and S etc.
F(n) is the diagonal sum of the entries in Pascal's triangle at 45 degrees slope. - , Dec 29 2001
F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(), May 19 2002
F(n+1) = number of ()-, ()- and ()-avoiding involutions in S_n.
This is also the Horadam sequence (0,1,1,1). - , Aug 18 2003
An INVERT transform of . INVERT([1,1,2,3,5,8,...]) gives . INVERT([1,2,3,5,8,13,21,...]) gives . - , Dec 12 2003
Number of meaningful differential operations of the k-th order on the space R^3. - , Mar 02 2004
F(n)=number of compositions of n-1 with no part greater than 2. Example: F(4)=3 because we have 3 = 1+1+1 = 1+2 = 2+1.
F(n) = number of compositions e.g., F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - , Jun 22 2004
F(n) = number of binary words of length n beginning with 0 and havin e.g., F(6) counts , , 0001. - , Jun 22 2004
The number of sequences (s(0),s(1),...,s(n)) such that 0&s(i)&5, |s(i)-s(i-1)|=1 and s(0)=1 is F(n+1); e.g., F(5+1) = 8 corresponds to , , 3434. - , Jun 22 2004 [corrected by Neven Juric, Jan 09 2009]
Likewise F(6+1) = 13 corresponds to these thirteen sequences with seven numbers: 11, 11, 33, 33, 1234343. - Neven Juric, Jan 09 2008
A relationship between F(n) and the Mandelbrot set is discussed in the link &Le nombre d'or dans l'ensemble de Mandelbrot& (in French). - , Sep 19 2004
For n&0, the continued fraction for F(2n-1)*Phi=[F(2n);L(2n-1),L(2n-1),L(2n-1),...] and the continued fraction for F(2n)*Phi=[F(2n+1)-1;1,L(2n)-2,1,L(2n)-2,...]. Also true: F(2n)*Phi=[F(2n+1);-L(2n),L(2n),-L(2n),L(2n),...] where L(i) is the i-th Lucas number ().... - , Nov 28 2004 [corrected by , Oct 20 2010]
F(n+1) (for n&=1) = number of permutations p of 1,2,3,...,n such that |k-p(k)|&=1 for k=1,2,...,n. (For &=2 and &=3, see
and .) - , Nov 28 2004
The ratios F(n+1)/F(n) for n&0 are the convergents to the simple continued fraction expansion of the golden section. - , Dec 19 2004
Lengths of successive words (starting with a) under the substitution: {a -& ab, b -& a}. - , Jan 22 2005
The Fibonacci sequence, like any additive sequence, naturally tends to be geometric with common ratio not a rational power of 10; consequently, for a sufficiently large number of terms, Benford's law of first significant digit (i.e., first digit 1 &= d &= 9 occurring with probability log_10(d+1) - log_10(d)) holds. - , Apr 29 2005
a(n) = Sum(abs((n, k)): 0 &= k &= n). - , Jun 01 2005
a(n) = ((n)).
Fib(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of . - , Mar 11 2003
Number of order ideals of the &zig-zag& poset. See vol. 1, ch. 3, prob. 23 of Stanley. - , Dec 27 2005
F(n+1)/F(n) is also the Farey fraction sequence (see
for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - , May 08 2006
a(n+2) is the number of paths through 2 plates of glass with n reflections (reflections occurring at plate/plate or plate/air interfaces). Cf. -. - , Jul 06 2006
F(n+1) equals the number of downsets (i.e., decreasing subsets) of an n-element fence, i.e., an ordered set of height 1 on {1,2,...,n} with 1 & 2 & 3 & 4 & ... n and no other comparabilities. Alternatively, F(n+1) equals the number of subsets A of {1,2,...,n} with the property that, if an odd k is in A, then the adjacent elements of {1,2,...,n} belong to A, i.e., both k - 1 and k + 1 are in A (provided they are in {1,2,...,n}). - , Aug 25 2006
Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - , Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, round(log_phi(sqrt((sqrt(5) a(n) + sqrt(5 a(n)^2 - 4))(sqrt(5) a(n) + sqrt(5 a(n)^2 + 4)))/2)) = n for n &= 3, obtained by rounding the arithmetic mean of the inverses given in
and . - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
A result of Jacobi from 1848 states that every symmetric matrix over a p.i.d. is congruent to a triple-diagonal matrix. Consider the maximal number T(n) of summands in the determinant of an n X n triple-diagonal matrix. This is the same as the number of summands in such a determinant in which the main-, sub- and super-diagonal elements are all nonzero. By expanding on the first row we see that the sequence of T(n)'s is the Fibonacci sequence without the initial stammer on the 1's. - Larry Gerstein (gerstein(AT)math.ucsb.edu), Mar 30 2007
Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd. There is a similar representation for Lucas numbers (). Many Fibonacci formulas now easily follow from appropriate sinh and cosh formulas. For example: the de Moivre theorem (cosh(x)+sinh(x))^m=cosh(mx)+sinh(mx) produces L(n)^2+5F(n)^2=2L(2n) and L(n)F(n)=F(2n) (setting x=n*psi and m=2). - , Apr 18 2007
Inverse: floor(log_phi(sqrt(5)*Fib(n))+1/2)=n, for n&1. Also for n&0, floor(1/2*log_phi(5*Fib(n)*Fib(n+1)))=n. Extension valid for integer n, except n=0,-1: floor(1/2*sign(Fib(n)*Fib(n+1))*log_phi|5*Fib(n)*Fib(n+1)|)=n (where sign(x) = sign of x). - , May 02 2007
F(n+2) = The number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007
From Kauffman and Lopes, Proposition 8.2, p. 21: &The sequence of the determinants of the Fibonacci sequence of rational knots is the Fibonacci sequence (of numbers).& - , Oct 26 2007
This is a_1(n) in the Doroslovacki reference.
Let phi = (sqrt(5)+1)/2 = 1.6180339...; then phi^n = (1/phi)*a(n) + a(n+1). Example: phi^4 = 6.8541019... = (0.6180339...)*3 + 5. Also phi = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + 1/(34*89) + ... - , Dec 15 2007
The sequence of first differences, Fib(n+1)-Fib(n), is essentially the same sequence: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... - , Mar 03 2008
a(n)= the number of different ways to run up a staircase with n steps, taking steps of odd sizes where the order is relevant and there is no other restriction on the number or the size of each step taken. - , May 21 2008
Equals row sums of triangle . - , Sep 12 2008
Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - , Sep 15 2008
F(n) is the number of possible binary sequences of length n that obey the sequential construction rule: if last symbol is 0, add the complement (1); else add 0 or 1. Here 0,1 are metasymbols for any 2-valued symbol set. This rule has obvious similarities to JFJ Laros's rule, but is based on addition rather than substitution and creates a tree rather than a single sequence. - , Oct 05 2008
F(n) = Product_{k=1..(n-1)/2} (1 + 4*cos^2 k*Pi/n), where terms = roots to the Fibonacci product polynomials, . - , Nov 22 2008
Fp == 5^((p-1)/2) mod p, p = prime [Schroeder, p. 90]. -
& , Feb 21 2009
(Ln)^2 - 5*(Fn)^2 = 4*(-1)^n. Example: 11^2 - 5*5 = -4. - , Mar 11 2009
Output of Kasteleyn's formula for the number of perfect matchings of an m X n grid specializes to the Fibonacci sequence for m=2. - Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009
(Fib(n),Fib(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 06 2009
(Fib(n),Fib(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 08 2009
a(n+2)=((n)). - , Sep 26 2009
Difference between of number of closed walks of length n+1 from a node on a pentagon and number of walks of length n+1 between two adjacent nodes on a pentagon. - , Feb 10 2010
F(n+1) = number of Motzkin paths of length n having exactly one weak ascent. A Motzkin path of length n is a lattice path from (0,0) to (n,0) consisting of U=(1,1), D=(1,-1) and H=(1,0) steps and never going below the x-axis. A weak ascent in a Motzkin path is a maximal sequence of consecutive U and H steps. Example: a(5)=5 because we have (HHHH), (HHU)D, (HUH)D, (UHH)D, and (UU)DD (the unique weak ascent is shown see ). - , Mar 11 2010
(F(n-1) + F(n+1))^2 - 5F(n-2)*F(n+2) = 9*(-1)^n. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Mar 31 2010
From the Pinter and Ziegler reference's abstract: authors &show that essentially the Fibonacci sequence is the unique binary recurrence which contains infinitely many three-term arithmetic progressions. A criterion for general linear recurrences having infinitely many three-term arithmetic progressions is also given.& - , May 22 2010
F(n+1) = number of paths of length n starting at initial node on the path graph P_4. - , May 27 2010
F(k) = Number of cyclotomic polynomials in denominator of generating function for number of ways to place k nonattacking queens on an n X n board. - , Jun 07 2010
As n-& inf., (a(n)/a(n-1) - a(n-1)/a(n)) tends to 1.0. Example: a(12)/a(11) - a(11)/a(12) = 144/89 - 89/144 = 0..... - , Jul 16 2010
From , Oct 20 2010: (Start)
Fibonacci numbers are those numbers m such that m*phi is closer to an integer than k*phi for all k, 1&=k&m. More formally: a(0)=0, a(1)=1, a(2)=1, a(n+1)=minimal m&a(n) such that m*phi is closer to an integer than a(n)*phi.
For all numbers 1&=k&Fib(n), the inequality |k*phi-round(k*phi)| & |Fib(n)*phi-round(Fib(n)*phi)| holds.
Fib(n)*phi - round(Fib(n)*phi) = -((-phi)^(-n)), for n&1.
Fract(0.5+Fib(n)*phi) = 0.5 -(-phi)^(-n), for n&1.
Fract(Fib(n)*phi) = (1/2)*(1+(-1)^n)-(-phi)^(-n), n&1.
Inverse: n = -log_phi |0.5-fract(0.5+Fib(n)*phi)|.
F((n)*k) mod n = 0, for any integer k. - , Nov 27 2010
F(n+k)^2-F(n)^2 = F(k)*F(2n+k), for even k. - , Dec 04 2010
F(n+k)^2+F(n)^2 = F(k)*F(2n+k), for odd k. - , Dec 04 2010
F(n) = round(phi* F(n-1)) for n&1. - , Jan 13 2012
For n & 0: a(n) = length of n-th row in Wythoff array . - , Jan 26 2012
From , Feb 22 2012: (Start)
The number of free permutations of [n].
The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).
The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)
The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - , Apr 19 2012
Two conjectures: For n & 1, F(n+2)^2 mod F(n+1)^2 = F(n)*F(n+1) - (-1)^n. For n & 0, (F(2n) + F(2n+2))^2 = F(4n+3) + sum_{k = 2..2n}F(2k). - , May 06 2012
From , Jan 30 2014: (Start)
Proof of Ratushnyak's first conjecture: For n & 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2F(n+1)^2.
Consider: F(n+2)^2 - F(n)*F(n+1) - 2F(n+1)^2
&&&&&&&&&= F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)
&&&&&&&&&=(F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))
&&&&&&&&&= F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.
Proof of second conjecture: L(n) stands for Lucas number sequence from .
Consider the fact that
&&&&L(2n+1)^2 = L(4n+2) - 2
&&&(F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2
&&&(F(2n) + F(2n+2))^2 = sum{k = 2..2n, F(2k)} + F(4n+3).
The relationship: INVERT transform of (1,1,0,0,0,...) = (1, 2, 3, 5, 8,...), while the INVERT transform of (1,0,1,0,1,0,1,...) = (1, 1, 2, 3, 5, 8,...) is equivalent to: The numbers of compositions using parts 1 and 2 is equivalent to the numbers of compositions using parts == 1 mod 2 (i.e., the odd integers). Generally, the numbers of compositions using parts 1 and k is equivalent to the numbers of compositions of (n+1) using parts 1 mod k. Cf.
for k = 3 and
for k = 4. Example: for k = 2, n = 4 we have the compositions (22; 211, 121; 112; 1111) = 5; but using parts 1 and 3 we have for n = 5: (311, 131, 113, 11111, 5) = 5. - , Jul 05 2012
The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)*F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^n*F(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called &delta-Fibonacci& numbers as defined in comments to
(see also the papers of Witula et al.). - , Jul 24 2012
F(n) is the number of different (n-1)-digit binary numbers such that all substrings of length & 1 have at least one digit equal to 1. Example: for n = 5 there are 8 binary numbers with n - 1 = 4 digits (, , , ), only the F(n) = 5 numbers ,
and 1111 have the desired property. - , Nov 30 2012
For positive n, F(n+1) equals the determinant of the n X n tridiagonal matrix with 1's along the main diagonal, i's along the superdiagonal and along the subdiagonal where i = sqrt(-1). Example: Det([1,i,0,0; i,1,i,0; 0,i,1,i; 0,0,i,1]) = F(4+1) = 5. - , Feb 24 2013
For n&=1, number of compositions of n where there is a drop between every second pair of parts, starting with the f see example. Also, a(n+1) is the number of compositions where there is a drop between every second pair of parts, starting with the s see example. - , May 21 2013
Central terms of triangles in
and , n & 0. - , Jul 28 2013
For n&=4, F(n-1) is the number of simple permutations in the geometric grid class given in . - , Sep 08 2013
a(n) are the pentagon (not pentagonal) numbers because the algebraic degree 2 number rho(5) = 2*cos(Pi/5) = phi (golden section), the length ratio diagonal/side in a pentagon, has minimal polynomial C(5,x) = x^2 - x - 1 (see , n=5), hence rho(5)^n = a(n-1)*1 + a(n)*rho(5), n &= 0, in the power basis of the algebraic number field Q(rho(5)). One needs a(-1) = 1 here. See also the P. Steinbach reference under . - , Oct 01 2013
(a(n)) = 1. - , Oct 10 2013
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n+2k)^2 - F(n)^2 = F(n+k)*( F(n+3k) - F(n-k) ). - , Dec 20 2013
( F(n), F(n+2k) ) satisfies the Diophantine equation: X^2 + Y^2 - L(2k)*X*Y = F(4k)^2*(-1)^n. This generalizes Bouhamida’s comments dated Sep 06 2009 and Sep 08 2009. - , Jan 07 2014
For any prime p there is an infinite periodic subsequence within F(n) divisible by p, that begins at index n = 0 with value 0, and its first nonzero term at n = (i), and period k = (i). Also see . - , Jan 26 2014
Range of row n of the circular Pascal array of order 5. - , May 30 2014 [orig. Kicey-Klimko 2011, and observations by Glen W more general work found in Ault-Kicey 2014]
Nonnegative range of the quintic polynomial 2*y - y^5 + 2*x*y^4 + x^2*y^3 - 2*x^3*y^2 - x^4*y with x, y &= 0, see Jones 1975. - , Jun 01 2014
The expression round(1/(F(k+1)/F(n) + F(k)/F(n+1))), for n & 0, yields a Fibonacci sequence with k-1 leading zeros (with rounding 0.5 to 0). - , Aug 04 2014
Conjecture: For n & 0, F(n) is the number of all admissible residue classes for which specific finite subsequences of the Collatz 3n + 1 function consists of n+2 terms. This has been verified for 0 & n & 51. For details see Links. - , Oct 03 2014
a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - , Oct 02 2014
a(1)=1=a(2), a(3)=2 are the only Fibonacci numbers that are of the form prime-1. - , Jun 07 2015
Any consecutive pair (m, k) of the Fibonacci sequence a(n) illustrates a fair equivalence between m miles and k kilometers. For instance, 8 miles ~ 13 13 miles ~ 21 km. -, Oct 06 2014
(n -& oo) lim (log F(n+1)/log F(n))^n = e. - , Oct 06 2014
a(n+1) counts closed walks on K_2, containing one loop on the other vertex. Equivalently the (1,1)_entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1; 1,1). - , Oct 29 2014
a(n-1) counts closed walks on the graph G(1-l-loop,2-loop). - , Nov 26 2014
From , Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers , with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers,
with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted , the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. ).
BTC(x) = C[Pinv(x)] gives , a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence , so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for , C[P[x,1-t]], and that for , Pinv[Cinv(x),t+1].
In keeping with historical accounts (see the references by P. Singh and S. Kak), the generalized Fibonacci sequence a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, ... can also be described as the Gopala-Hemachandra numbers H(n) = H(n-1) + H(n-2), with F(n) = H(n) for a = b = 1, and Lucas sequence L(n) = H(n) for a = 2, b = 1. - , Jan 11 2015
D. E. Knuth writes: &Before Fibonacci wrote his work, the sequence F_{n} had already been discussed by Indian scholars, who had long been interested in rhythmic patterns that are formed from one-beat and two-beat notes. The number of such rhythms having n beats altogether is F_{n+1}; therefore both Gopāla (before 1135) and Hemachandra (c. 1150) mentioned the numbers 1, 2, 3, 5, 8, 13, 21, ... explicitly.& (TAOCP Vol. 1, 2nd ed.) - , Jan 11 2015
F(n+1) equals the number of binary words of length n avoiding runs of zeroes of odd lengths. - , Jan 28 2015
From , Apr 12 2015: (Start)
We prove Conjecture 1 of Rashid listed in the Formula section.
We use the following notation: F(n)=(n), the Fibonacci numbers, and L(n) = (n), the Lucas numbers. The fundamental Fibonacci-Lucas recursion asserts that G(n) = G(n-1)+ G(n-2), with &L& or &F& replacing &G&.
We need the following prerequisites which we label (A), (B),(C), (D). The prerequisites are formulas in the Koshy book listed in the References section. (A) F(m-1)+F(m+1) = L(m) (Koshy, p. 97, #32), (B) L(2m)+2(-1)^m = L(m)^2 (Koshy p. 97, #41), (C) F(m+k)F(m-k) = (-1)^n F(k)^2 (Koshy, p. 113, #24, Tagiuri's identity), and (D) F(n)^2+F(n+1)^2 = F(2n+1) (Koshy, p. 97, #30).
We must also prove (E), L(n+2) F(n-1) = F(2n+1)+2(-1)^n. To prove (E), first note that by (A), proof of (E) is equivalent to proving that F(n+1)F(n-1) + F(n+3)F(n-1) = F(2n+1)+2(-1)^n. But by (C) with k=1, we have F(n+1)F(n-1) = F(n)^2 +(-1)^n. Applying (C) again with k=2 and m=n+1, we have F(n+3)F(n-1) = F(n+1)+(-1)^n. Adding these two applications of (C) together and using (D) we have, F(n+1)F(n-1) + F(n+3)F(n-1) = F(n)^2 + F(n+1)^2 + 2(-1)^n = F(2n+1)+2(-1)^n, completing the proof of (E).
We now prove Conjecture 1. By (A) and the Fibonacci-Lucas recursion, we have F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4) = [F(2n+1)+F(2n+3)] + [F(2n+2)+F(2n+4)] = L(2n+2)+L(2n+3)=L(2n+4). But then by (B), with m=2n+4, we have sqrt(L(2n+4)+2(-1)^n)) = L(n+2). Finally by (E), we have L(n+2) F(n-1)= F(2n+1)+2*(-1)^n. Dividing both sides by F(n-1), we have (F(2n+1)+2*(-1)^n)/F(n-1) = L(n+2) = sqrt(F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4)+2(-1)^n), as required.
In Fibonacci's Liber Abaci the rabbit problem appears in the translation of L. E. Sigler on pp. 404-405, and a remark [27] on p. 637. - , Apr 17 2015
a(n) counts partially ordered partitions of (n-1) into parts 1,2,3 where only the order of adjacent 1's and 2's are unimportant. (See example.) - , Jul 27 2015
F(n) divides F(nk). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - , Aug 24 2015
F(n) is the number of UDU-equivalence classes of ballot paths of length n. Two ballot paths of length n with steps U = (1,1), D = (1,-1) are UDU-equivalent whenever the positions of UDU are the same in both paths. - , Aug 25 2015
Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1 is the basis for a geometrical paradox (or dissection fallacy) in . - , Oct 23 2015
For n &= 4, F(n) is the number of up-down words on alphabet {1,2,3} of length n-2. - , Nov 23 2015
F(n+2) is the number of terms in p(n), where p(n)/q(n) is the n-th convergent of the formal infinite continued fraction [a(0),a(1),...]; e.g., p(3) = a(0)a(1)a(2)a(3) + a(0)a(1) + a(0)a(3) + a(2)a(3) + 1 has F(5) terms. Also, F(n+1) is the number of terms in q(n). - , Dec 23 2015
F(n+1) (for n&=1) is the permanent of an n X n matrix M with M(i,j)=1 if |i-j|&=1 and 0 otherwise. - , Jan 08 2016
A trapezoid has three sides of lengths in order F(n), F(n+2), F(n). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*F(n+1). For a trapezoid with lengths of sides in order F(n+2), F(n), F(n+2), the fourth side will be F(n+3). - , Mar 17 2016
(1) Join two triangles with lengths of sides L(n), F(n+3), L(n+2) and F(n+2), L(n+1), L(n+2) (where L(n)=(n)) along the common side of length L(n+2) to create an irregular quadrilateral. Its area is approximately (5*F(2*n-1) - (F(2*n-7) - F(2*n-13))/5. (2) Join two triangles with lengths of sides L(n), F(n+2), F(n+3) and L(n+1), F(n+1, F(n+3) along the common side F(n+3) to form an irregular quadrilateral. Its area is approximately 4*F(2*n-1) - 2*(F(2*n-7) + F(2*n-18)). - , Apr 06 2016
From , Jun 13 2016: (Start)
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2, x}, g(3) = {3, 2x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If a positive integer N is not a square and r = sqrt(N), then the number of (not necessarily distinct) integers in g(n) is (n), for n & = 1. See . (End)
Consider the partitions of n, with all summands initially listed in nonincreasing order. Freeze all the 1's in place and then allow all the other summands to change their order, without displacing any of the 1's. The resulting number of arrangements is a(n+1). - , Jun 14 2016
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, signature (1,1).
G.f.: x / (1 - x - x^2).
G.f.: Sum_{n&=0} x^n * Product_{k=1..n} (k + x)/(1 + k*x). - , Oct 26 2013
F(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5)).
Alternatively, F(n) = ((1/2+sqrt(5)/2)^n-(1/2-sqrt(5)/2)^n)/sqrt(5).
F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n).
F(n) = round(phi^n/sqrt(5)).
F(n+1) = Sum_{j=0..[n/2]} binomial(n-j, j).
This is a d that is, if n divides m, then a(n) divides a(m). - , Apr 07 2012
E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2). - , Nov 30 2001
[0 1; 1 1]^n [0 1] = [F(n); F(n+1)]
x | F(n) ==& x | F(kn).
A sufficient condition for F(m) to be divisible by a prime p is (p - 1) divides m, if p == 1 or 4 (mod 5); (p + 1) divides m, if p == 2 or 3 (mod 5); or 5 divides m, if p = 5. (This is essentially Theorem 180 in Hardy and Wright.) - Fred W. Helenius (fredh(AT)), Jun 29 2001
a(n)=F(n) has the property: F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1). - , Nov 13 2003
, Feb 21 2004, makes 4 conjectures and gives 3 theorems:
Conjecture 1: for n&=2 sqrt{F(2n+1)+F(2n+2)+F(2n+3)+F(2n+4)+2*(-1)^n}={F(2n+1)+2*(-1)^n}/F(n-1). [For a proof see Comments section.]
Conjecture 2: for n&=0, {F(n+2)* F(n+3)}-{F(n+1)* F(n+4)}+ (-1)^n = 0.
Conjecture 3: for n&=0, F(2n+1)^3 - F(2n+1)*[(2*A^2)-1] - [A + A^3]=0, where A = {F(2n+1)+sqrt{5*F(2n+1)^2 +4}}/2.
Conjecture 4: for x&=5, if x is a Fibonacci number &= 5 then g*x*[{x+sqrt{5*(x^2) +- 4}}/2]*[2x+{{x+sqrt{5*(x^2) +- 4}}/2}]*[2x+{{3x+3*sqrt {5*(x^2) +- 4}}/2}]^2+[2x+{{x+sqrt{5*(x^2) +- 4}}/2}] +- x*[2x+{{3x+3*sqrt{5*(x^2) +- 4}}/2}]^2 -x*[2x+{{x+sqrt{5*(x^2) +- 4}}/2}]*[x+{{x+sqrt{5*(x^2) +- 4}}/2}]* [2x+{{3x+3*sqrt{5*(x^2) +- 4}}/2}]^2= 0, where g = {1 + sqrt(5)/2}.
Theorem 1: for n&=0, {F(n+3)^ 2 - F(n+1)^ 2}/F(n+2)={F(n+3)+ F(n+1)}.
Theorem 2: for n&=0, F(n+10) = 11*F(n+5) + F(n).
Theorem 3: for n&=6, F(n) = 4*F(n-3) + F(n-6).
Conjecture 2 of Rashid is actually a special case of the general law F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1) (take n &- n+1 and m &- -(n+4) in this law). - Harmel Nestra (harmel.nestra(AT)ut.ee), Apr 22 2005
Conjecture 2 of Rashid Kurmang simplified: F(n)*F(n+3) = F(n+1)*F(n+2)-(-1)^n. Follows from d'Ocagne's identity: m=n+2. - , May 06 2012
Conjecture: for all c such that 2-Phi &= c & 2*(2-Phi) we have F(n) = floor(Phi*a(n-1)+c) for n & 2. - , Jul 21 2004
|2*Fib(n) - 9*Fib(n+1)| = 4*(n) + (n+1). - , Aug 13 2004
For x & Phi, Sum_{n=0..inf} F(n)/x^n = x/(x^2 - x - 1) - , Oct 27 2004
F(n+1) = exponent of the n-th term in the series f(x, 1) determined by the equation f(x, y) = xy + f(xy, x). - , Dec 19 2004
a(n-1) = Sum_{k=0..n} (-1)^k*binomial(n-ceil(k/2), floor(k/2)). - , May 05 2005
F(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)(1+(-1)^(n-k))/2. - , Aug 28 2005
Fibonacci(n) = Product(1 + 4[cos(j*Pi/n)]^2, j=1..ceil(n/2)-1). [Bicknell and Hoggatt, pp. 47-48.] - , Oct 15 2006
F(n) = 2^-(n-1)*Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)*5^k. - , Feb 07 2006
a(n) = (b(n+1)+b(n-1))/n where {b(n)} is the sequence . - , Nov 22 2006
F(n*m) = Sum_{k = 0..m} binomial(m,k)*F(n-1)^k*F(n)^(m-k)*F(m-k). The generating function of F(n*m) (n fixed, m = 0,1,2...) is G(x) = F(n)*x / ((1-F (n-1)*x)^2-F(n)*x*(1-F(n-1)*x)-( F(n)*x)^2). E.g., F(15) = 610 = F(5*3) = binomial(3,0)* F(4)^0*F(5)^3*F(3) + binomial(3,1)* F(4)^1*F(5)^2*F(2) + binomial(3,2)* F(4)^2*F(5)^1*F(1) + binomial(3,3)* F(4)^3*F(5)^0*F(0) = 1*1*125*2 + 3*3*25*1 + 3*9*5*1 + 1*27*1*0 = 250 + 225 + 135 + 0 = 610. - , Feb 12 2007
From , Mar 19 2007: (Start)
&&Let L(n) = (n) = Lucas numbers. Then:
&&For a&=b and odd b, F(a+b)+F(a-b)=L(a)*F(b).
&&For a&=b and even b, F(a+b)+F(a-b)=F(a)*L(b).
&&For a&=b and odd b, F(a+b)-F(a-b)=F(a)*L(b).
&&For a&=b and even b, F(a+b)-F(a-b)=L(a)*F(b).
&&F(n+m)+(-1)^m*F(n-m)=F(n)*L(m);
&&F(n+m)-(-1)^m*F(n-m)=L(n)*F(m);
&&F(n+m+k)+(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k)) =F(n)*L(m)*L(k);
&&F(n+m+k)-(-1)^k*F(n+m-k)+(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k)) =L(n)*L(m)*F(k);
&&F(n+m+k)+(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)+(-1)^k*F(n-m-k)) =L(n)*F(m)*L(k);
&&F(n+m+k)-(-1)^k*F(n+m-k)-(-1)^m*(F(n-m+k)-(-1)^k*F(n-m-k)) =5*F(n)*F(m)*F(k). (End)
A corollary to Kristof 2007 is 2*F(a+b)=F(a)*L(b)+L(a)*F(b). - , Apr 24 2014
For n&m, the sum of the 2m consecutive Fibonacci numbers F(n-m-1) thru F(n+m-2) is F(n)*L(m) if m is odd, and L(n)*F(m) if m is even (see the McRae link). - , Apr 24 2014.
Fib(n) = b(n)+(p-1)*Sum_{1&k&n} floor(b(k)/p)*Fib(n-k+1) where b(k) is the digital sum analog of the Fibonacci recurrence, defined by b(k)=ds_p(b(k-1))+ds_p(b(k-2)), b(0)=0, b(1)=1, ds_p=digital sum base p. Example for base p=10: Fib(n)=(n)+9*Sum_{1&k&n} ((k))*Fib(n-k+1). - , Jul 01 2007
Fib(n) = b(n)+p*Sum_{1&k&n} floor(b(k)/p)*Fib(n-k+1) where b(k) is the digital product analog of the Fibonacci recurrence, defined by b(k)=dp_p(b(k-1))+dp_p(b(k-2)), b(0)=0, b(1)=1, dp_p=digital product base p. Example for base p=10: Fib(n)=(n)+10*Sum_{1&k&n} ((k))*Fib(n-k+1). - , Jul 01 2007
a(n) = denominator of continued fraction [1,1,1,...] (with n ones); e.g., 2/3 = continued fraction [1,1,1]; where barover[1] = [1,1,1...] = 0.6180339.... - , Nov 29 2007
F(n + 3) = 2F(n + 2) - F(n), F(n + 4) = 3F(n + 2) - F(n), F(n + 8) = 7F(n + 4) - F(n), F(n + 12) = 18F(n + 6) - F(n). - , Feb 01 2008
1 = 1/(1*2) + 1/(1*3) + 1/(2*5) + 1/(3*8) + 1/(5*13) + ... = 1/2 + 1/3 + 1/10 + 1/24 + 1/65 + 1/168 + ...; where
= (0, 2, 3, 10, 24, 65, 168,...). - , Mar 16 2008
a(2^n) = prod{i=0}^{n-2}B(i) where B(i) is . Example 3*7*47 = Fib(16). - , Apr 23 2008
F(n) = (1/(n-1)!) * (n^(n-1) - (C(n-2,0) + 4*C(n-2,1) + 3*C(n-2,2))*n^(n-2) + (10*C(n-3,0) + 49*C(n-3,1) + 95*C(n-3,2) + 83*C(n-3,3) + 27*C(n-3,4))*n^(n-3) - (90*C(n-4,0) + 740*C(n-4,1) + 2415*C(n-4,2) + 4110*C(n-4,3) + 3890*C(n-4,4) + 1950*C(n-4,5) + 405*C(n-4,6))*n^(n-4) + ... ). - , Nov 24 2004
a(n+1) = Sum_{k, 0&=k&=n} (n,k)*(-1)^(n-k). -, Oct 26 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}... Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n), where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i + l_(i+1) &= 2 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - , Feb 25 2009
a(n+1) = 2^n sqrt(Product_{k=1..n} cos(k Pi/(n+1))^2+1/4)) (Kasteleyn's formula specialized). - Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009
a(n+1) = Sum_{k=floor[n/2] mod 5} C(n,k) - Sum_{k=floor[(n+5)/2] mod 5} C(n,k) = (n) - (n) = |(n)-(n-1)|. - , Feb 10 2010
If p[i]=modp(i,2) and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i&=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n&=1, a(n)=det A. - , May 02 2010
Limit(F(k+n)/F(k), k = infinity) = (L(n) + F(n)*sqrt(5))/2 with the Lucas numbers L(n)= (n). - , May 27 2010
For n&=1, F(n)=round(log_2(2^{phi*F(n-1)} + 2^{phi*F(n-2)})), where phi is the golden ratio. - , Jun 24 2010, Jun 27 2010
For n&=1, a(n+1)=ceil(phi*a(n)), if n is even and a(n+1)=floor(phi*a(n)), if n is odd (phi = golden ratio). - , Jul 01 2010
a(n) = 2*a(n-2) + a(n-3), n&2. - , Sep 08 2010
a(2^n) = Prod_{i=0..n-1} (2^i). - , Nov 28 2010
a(n)^2 - a(n-1)^2 = a(n+1)*a(n-2), see .
a(n) = sqrt((-1)^k*(a(n+k)^2 - a(k)*a(2n+k))), for any k. - , Dec 03 2010
F(2*n) = F(n+2)^2 - F(n+1)^2 - 2*F(n)^2. - , Jun 04 2011
(-1)^(n+1) = F(n)^2 + F(n)*F(1+n) - F(1+n)^2.
&&F(n) = -F(n+2)(-2 + (F(n+1))^4 + 2*(F(n+1)^3*F(n+2)) - (F(n+1)*F(n+2))^2 2*F(n+1)(F(n+2))^3 + (F(n+2))^4)- F(n+1). - , Nov 17 2011
F(n) = 1 + Sum_{x=1..n-2} F(x). - , Feb 05 2012
F(n) = 4*F(n-2) - 2*F(n-3) - F(n-6). - , Apr 01 2012
F(n) = round(phi^(n+1)/(phi+2)). - , Apr 20 2012
From , Jun 03 2012: (Start)
G.f. A(x) = x/(1-x-x^2) = G(0)/sqrt(5) where G(k)= 1 -((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*c^k/G(k+1))) and a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
Let E(x) be the e.g.f., i.e.,
E(x) = 1*x + 1/2*x^2 + 1/3*x^3 + 1/8*x^4 + 1/24*x^5 + 1/90*x^6 + 13/5040*x^7 + ...; then
E(x) = G(0)/sqrt(5); G(k)= 1 -((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*(k+1)*c^k/G(k+1))), where a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
From , Nov 30 2012: (Start)
Fib(n) = 1 + Sum_{j_1=1..n-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} 1 + ... + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} ... Sum_{j_k=1..j_(k-1)-2} 1, where k = floor((n-1)/2).
Example: Fib(6) = 1 + Sum_{j=1..4} 1 + Sum_{j=1..4} Sum_{k=1..(j-2)} 1 + 0 = 1 + (1 + 1 + 1 + 1) + (1 + (1 + 1)) = 8.
Fib(n) = Sum_{j=0..k} S(j+1,n-2j), where k = floor((n-1)/2) and the S(j,n) are the n-th j-simplex sums: S(1,n) = 1 is the 1-simplex sum, S(2,n) = Sum_{k=1..n} S(1,k) = 1+1+...+1 = n is the 2-simplex sum, S(3,n) = Sum_{k=1..n} S(2,k) = 1+2+3+...+n is the 3-simplex sum (= triangular numbers = ), S(4,n) = Sum_{k=1..n} S(3,k) = 1+3+6+...+n(n+1)/2 is the 4-simplex sum (= tetrahedral numbers = ) and so on.
Since S(j,n) = binomial(n-2+j,j-1), the formula above equals the well-known binomial formula, essentially. (End)
G.f. A(x) = x / (1 - x / (1 - x / (1 + x))). - , Jan 04 2013
Sum_{n&=1} (-1)^(n-1)/(a(n)*a(n+1)) = 1/phi (phi=golden ratio). - , Feb 22 2013
From , Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (a(n) + sqrt(5*(a(n))^2 + 4*(-1)^n))/2;
(2) Sum_{k=1...n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(3) a(n)/a(n+1) = 1/phi + r(n), where |r(n)| & 1/(a(n+1)*a(n+2)). (End)
F(n+1) = F(n)/2 + sqrt((-1)^n + 5*F(n)^2/4), n&=0. F(n+1) = U_n(i/2)/i^n, (U:= Chebyshef 2nd kind). - , Mar 04 2013
G.f.: -Q(0) where Q(k) = 1 - (1+x)/(1 - x/(x - 1/Q(k+1) )); (continued fraction). - , Mar 06 2013
G.f.: x-1-1/x + 1/x/Q(0), where Q(k) = 1 - (k+1)*x/(1 - x/(x - (k+1)/Q(k+1))); (continued fraction). - , Apr 23 2013
G.f.: x*G(0), where G(k)= 1 + x*(1+x)/(1 - x*(1+x)/(x*(1+x) + 1/G(k+1) )); (continued fraction). - , Jul 08 2013
G.f.: x^2 - 1 + 2*x^2/(W(0)-2), where W(k) = 1 + 1/(1 - x*(k + x)/( x*(k+1 + x) + 1/W(k+1) )); (continued fraction). - , Aug 28 2013
G.f.: Q(0) -1, where Q(k) = 1 + x^2 + (k+2)*x -x*(k+1 + x)/Q(k+1); (continued fraction). - , Oct 06 2013
Let b(n) = b(n-1) + b(n-2), with b(0) = 0, b(1) = phi. Then, for n&=2, F(n)= floor(b(n-1)) if n is even, F(n) = ceil(b(n-1)), if n is odd, with convergence. - , Jan 19 2014
a(n) = Sum_{t1*g(1)+t2*g(2)+...+tn*g(n)=n} multinomial(t1+t2 +...+tn,t1,t2,...,tn), where g(k)=2*k-1. - , Feb 27 2014
F(n) = round(sqrt(F(n-1)^2 + F(n)^2 + F(n+1)^2)/2), for n & 0. This rule appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - , Jul 27 2014
F(n) = round(2/(1/F(n) + 1/F(n+1) + 1/F(n+2)), for n & 0. This rule also appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - , Aug 03 2014
F(n) = round(1/(Sum_{j&=n+2} 1/F(j))). - , Aug 14 2014
a(n) = hypergeometric([-n/2+1/2, -n/2+1], [-n+1], -4) for n&=2. - , Sep 19 2014
F(n) = (L(n+1)^2 - L(n-1)^2)/(5*L(n)), where L(n) is (n), with a similar inverse relationship. - , Nov 17 2014
Consider the graph G[1-1-loop,2-loop] in comment above. Construct the power matrix array T(n,j)=[A^*j]*[S^*(j-1)] where A=(1,1,0,...) and S=(0,1,0,...)(). [* is convolution operation] Define S^*0=I with I=(1,0,...). Then T(n,j) counts n-walks containing (j) loops and a(n-1) = Sum_{j=1...n} T(n,j). - , Nov 21 2014
Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n) = F(k)*F(n-k+3) - F(k-1)*F(n-k+2) - F(k-2)*F(n-k) + (-1)^k*F(n-2k+2). - , Dec 04 2014
F(n+k)^2 - L(k)*F(n)*F(n+k) + (-1)^k*F(n)^2 = (-1)^n*F(k)^2, if L(k) = (k). - , Jul 20 2015
F(2*n) = F(n+1)^2 - F(n-1)^2, similar to Koshy (D) and Forberg 2011, but different. - , Aug 12 2015
F(n+1) = ceiling( (1/phi)*Sum_{k=0..n} F(k) ). - , Sep 10 2015
a(n) = (L(n-3) + L(n+3))/10 where L(n)=(n). - , Nov 25 2015
From , Mar 27 2016 (Start):
F(n) = (F(2n+k+1) - F(n+1)*F(n+k+1))/F(n+k), k&=0.
Thus when k=0: F(n) = sqrt(F(2n+1) - F(n+1)^2).
F(n) = cbrt(F(3n) - F(n+1)^3 + F(n-1)^3).
F(n+2k) = binomial transform of any subsequence starting with F(n). Example F(6)=8: 1*8 = F(6)=8; 1*8 + 1*13 = F(8)=21; 1*8 + 2*13 + 1*21 = F(10)=55; 1*8 + 3*13 + 3*21 + 1*34 = F(12)=144, etc. This formula applies to Fibonacci-type sequences with any two seed values for a(0) and a(1) (e.g., Lucas sequence : a(0)=2, a(1)=1).
For x = 0,1,2,3,4, x=1/(x+1) = 1, 1/2, 2/3, 3/5, 5/8. These fractions have numerators 1,1,2,3,5, which are the 2nd to 6th entries in the sequence. - , Sep 15 2008
From , May 21 2013: (Start)
There are a(7)=13 compositions of 7 where there is a drop between every second pair of parts, starting with the first and second part:
01:&&[ 2 1 2 1 1 ]
02:&&[ 2 1 3 1 ]
03:&&[ 2 1 4 ]
04:&&[ 3 1 2 1 ]
05:&&[ 3 1 3 ]
06:&&[ 3 2 2 ]
07:&&[ 4 1 2 ]
08:&&[ 4 2 1 ]
09:&&[ 4 3 ]
10:&&[ 5 1 1 ]
11:&&[ 5 2 ]
12:&&[ 6 1 ]
13:&&[ 7 ]
There are abs(a(6+1))=13 compositions of 6 where there is no rise between every second pair of parts, starting with the second and third part:
01:&&[ 1 2 1 2 ]
02:&&[ 1 3 1 1 ]
03:&&[ 1 3 2 ]
04:&&[ 1 4 1 ]
05:&&[ 1 5 ]
06:&&[ 2 2 1 1 ]
07:&&[ 2 3 1 ]
08:&&[ 2 4 ]
09:&&[ 3 2 1 ]
10:&&[ 3 3 ]
11:&&[ 4 2 ]
12:&&[ 5 1 ]
13:&&[ 6 ]
Partially ordered partitions of (n-1) into parts 1,2,3 where only the order of the adjacent 1's and 2's are unimportant. E.g., a(8)=21. These are (331),(313),(133),(322),(232),(223),(3211),(2311),(1321),(2131),(1132),(2113),(31111),(13111),(11311),(11131),(11113),(2221),(22111),(211111),(1111111). - , Jul 25 2015
Consider the partitions of 7 with summands initially listed in nonincreasing order. Keep the 1's frozen in position,(indicated by &[]&) and then allow the other summands to otherwise vary their order: 7; 6,[1]; 5,2; 2,5; 4,3; 3,4; 5,[1,1], 4,2,[1]; 2,4,[1]; 3,3,[1]; 3,3,2; 3,2,3; 2,3,3; 4,[1,1,1]; 3,2,[1,1]; 2,3,[1,1]; 2,2,2,[1]; 3,[1,1,1,1]; 2,2,[1,1,1]; 2,[1,1,1,1,1]; [1,1,1,1,1,1,1]. There are 21 = a(7+1) arrangements in all. - , Jun 14 2016
:= proc(n) combinat[fibonacci](n);
ZL:=[S, {a = Atom, b = Atom, S = Prod(X, Sequence(Prod(X, b))), X = Sequence(b, card &= 1)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=0..38); # , Apr 04 2008
spec := [B, {B=Sequence(Set(Z, card&1))}, unlabeled ]: seq(combstruct[count](spec, size=n), n=1..39); # , Apr 04 2008
# The following Maple command isFib(n) yields true or false depending on whether n is a Fibonacci number or not.
with(combinat): isFib := proc(n) local a: a := proc(n) local j: for j while fibonacci(j) &= n do fibonacci(j) end do: fibonacci(j-1) end proc: evalb(a(n) = n) end proc: # , Nov 11 2014
MATHEMATICA
Table[ Fibonacci[ k ], {k, 0, 50} ] (* to generate 0 as well as all other nonzero Fibonacci numbers, the lower limit is changed from 1 to 0; , Jul 11 2015 *)
Table[ 2^n Sqrt@Product[( Cos[Pi k/(n + 1)]^2 + 1/4), {k, n}]//FullSimplify, {n, 15}]; (* Kasteleyn's formula specialized, Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009 *)
Table[Fibonacci[n]^5 - Fibonacci[1 + n] + 3 Fibonacci[n]^4 Fibonacci[1 + n] + Fibonacci[n]^3 Fibonacci[1 + n]^2 - 3 Fibonacci[n]^2 Fibonacci[1 + n]^3 - Fibonacci[n] Fibonacci[1 + n]^4 + Fibonacci[1 + n]^5, {n, 1, 10}] (* , Nov 17 2011 *)
LinearRecurrence[{1, 1}, {0, 1}, 40] (* , Aug 03 2014 *)
(Axiom) [fibonacci(n) for n in 0..50]
(MAGMA) [Fibonacci(n): n in [0..38]];
(Maxima) makelist(fib(n), n, 0, 100); /* , Oct 21 2012 */
(PARI) a(n) = fibonacci(n)
(PARI) a(n) = imag(quadgen(5)^n)
(PARI) a(n)=my(phi=quadgen(5)); (phi^n-(-1/phi)^n)/(2*phi-1) \\ , Jun 17 2012
(PARI) a(n)=polcoeff(sum(m=0, n, x^m*prod(k=1, m, k+x +x*O(x^n))/prod(k=1, m, 1+k*x +x*O(x^n))), n) \\ , Oct 26 2013
(Python) # , Jan 05 2007 (Change leading dots to blanks.)
def fib():
... &&& Generates the Fibonacci numbers, starting with 0 &&&
... x, y = 0, 1
... while 1:
....... yield x
....... x, y = y, x+y
a = [f.next() for i in range(100)]
... &&& Returns Fibonacci number with index n, offset 0, 4 &&&
... return a[n]
................
def _list(N):
... &&& Returns a list of the first n Fibonacci numbers &&&
... return a[:N]
(Sage) ## Demonstration program from Jaap Spies:
a = sloane.; ## choose sequence
print a ## This returns the name of the sequence.
print a(38) ## This returns the 38th number of the sequence.
print a.list(39) ## This returns a list of the first 39 numbers.
-- Based on code from http://www.haskell.org/haskellwiki/The_Fibonacci_sequence
-- which also has other versions.
fib :: Int -& Integer
fib n = fibs !! n
.... fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
{- Example of use: map fib [0..38] , Sep 29 2009 -}
(Sage) [i for i in fibonacci_sequence(0, 40)] # , Jun 26 2014
(MAGMA) [0, 1] cat [n: n in [1..] | IsSquare(5*n^2-4) or IsSquare(5*n^2+4)]; // , Nov 19 2014
(signed Fibonacci numbers),
(complement), , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , .
First row of arrays , . Second row of arrays , , and
(k-generalized Fibonacci numbers).
a(n) = (4, n). a(n) = (0, j, n).
a(n) = (0, n+1) = (0, n+1) = (1, n-1) = (0, n) = (1, n-1), for n & 0.
Fibonacci-Pascal triangles: , , , , , , , , , .
Boustrophedon transforms: , .
Powers: , , .
Sequence in context:
Adjacent sequences:&&
core,nonn,nice,easy,,changed
, Apr 30 1991
Maintained by
Last modified August
4 18:06 EDT 2016.
Contains 275300 sequences.}
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