Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have new Fibonacci number That I want to know is there any special direct formula to count f(n).
like the normal Fibonacci:
F(0) = 7,
F(1) = 11,
F(n) = F(n-1) + F(n-2) (n>=2)
For example I want to count f(100) without knowing f(99) or f(98).

share|cite|improve this question
There is. I don't want to go looking for it right now, however. – Jan Dvorak Jan 11 '14 at 9:35
up vote 2 down vote accepted

Let $\mathcal{F}(x)$ be the ordinary generating function for the new Fibonacci number: $$\mathcal{F}(x)=\sum_{n\geq 1}F_n x^n$$ Note that \begin{align*} \mathcal{F}(x) = 7x+ 11x^2+ &18x^3+29x^4+47x^5+\cdots\\ x\cdot \mathcal{F}(x) =~~~~~~~~~~~ 7x^2+&11x^3+18x^4+29x^5+47x^6+\cdots\\ x^2\cdot \mathcal{F}(x) =~~~~~~~~~~~~~~~~~~~~~ &7x^3+11x^4+18x^5+29x^6+47x^7+\cdots \end{align*} Then $$(1-x-x^2)\mathcal{F}(x)=7x+4x^2$$ So, $$\mathcal{F}(x)=\frac{7x+4x^2}{1-x-x^2}=7\mathcal{F}_0(x)+4x\mathcal{F}_0(x)$$ where $\mathcal{F}_0(x)$ is the ordinary generating function for the standard Fibonacci numbers, i.e., $\mathcal{F}_0(x)=\sum_{n\geq 1}F^*_{n}x^n$ in which $F^*_n$ is the given by the well-known Binet's formula: $$F^*_n=\frac{(r)^n-(r')^n}{2^n\sqrt{5}}$$ where $r=1+\sqrt{5}$ and $r'=1-\sqrt{5}$

Now we are ready to express the coefficient of $x^n$ in $\mathcal{F}(x)$: $$F_n=7F^*_n+4F^*_{n-1}=7\frac{(r)^n-(r')^n}{2^n\sqrt{5}}+4\frac{(r)^{n-1}-(r')^{n-1}}{2^{n-1}\sqrt{5}}$$

share|cite|improve this answer

Since $(7,11) = (7,7)+(0,4)$, your sequence is $F(n)=4 Fibonacci(n) + 7 Fibonacci(n+1)$.

share|cite|improve this answer

The usual method to solve this is :



  1. Find the solutions of $X^2=X+1$. This are the roots $r_1$ and $r_2$ of the polynomial $X^2-X-1$. Hence $r_i=\frac{1\pm\sqrt{5}}{2}$
  2. As there are two distinct roots, then $$f_n=A.r_1^n+B.r_2^n$$
  3. Find $A$ and $B$ with the initial values of $f_n$. $$f_0=7=A+B$$ $$f_1=11=A.r_1+B.r_2$$ So $A=\frac{11-7.r_2}{r_1-r_2}$ and $B=\frac{11-7r_1}{r_2-r_1}$
share|cite|improve this answer

In general, for $\{f_n\} (n=0,1,\cdots)$ such that $f_{n}=f_{n-1}+f_{n-2}\ \ (n\ge 2)$, we have $$f_n=\frac{(\beta^n-\alpha^n)f_1-(\alpha\beta^n-\alpha^n\beta)f_0}{\beta-\alpha}$$ where $$\alpha=\frac{1-\sqrt 5}{2},\beta=\frac{1+\sqrt 5}{2}.$$

So, you can use this as $$f_{100}=\frac{(\beta^{100}-\alpha^{100})\times 11-(\alpha\beta^{100}-\alpha^{100}\beta)\times 7}{\beta-\alpha}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.