0
$\begingroup$

Denote $(f_n)_n$ the Fibonacci numbers and $(l_n)_n$ the Lucas numbers. The definition is such that $$ \begin{array}{lll} f_0 = 0,& f_1= 1,& f_{n+2}=f_{n+1}+f_n,\\ l_0 = 2,& l_1=1, & l_{n+2}=l_{n+1}+l_n. \end{array} $$

Define formal series $$F_{k}(x) := \sum_{n\geq 0} f_{k(n+1)}x^n \quad \text{and} \quad L_k(x) := \left\{ \begin{array}{ll} \frac{1}{1-x},& k=0, \\ \sum_{n\geq 0 } l_{k(n+1)}x^n, &k>0. \end{array}\right. $$

Prove that, for all $p\geq 0$:

\begin{eqnarray} \sum_{n\geq 0} f^{2p}_{n+1}x^n =& \frac{1}{5^p}\sum_{k=0}^p \binom{2p}{p-k}L_{2k}((-1)^{p-k}x) , \\ \\ \sum_{n\geq 0} f^{2p+1}_{n+1}x^n = &\frac{1}{5^p}\sum_{k=0}^p \binom{2p+1}{p-k}F_{2k+1}((-1)^{p-k}x) . \end{eqnarray}

My interest in that formula comes from the fact that $$ F_{2k+1}(x) = \frac{f_{2k+1}}{1-l_{2k+1}x-x^2}\quad \text{and} \quad L_{2k}(x) = \frac{l_{2k}-2x}{1-l_{2k}x+x^2}, $$ thus giving the generating series as a rational function, from which a recurrence relation for Fibonacci powers can be read off.

It is known that Fibonacci powers for a fixed exponent must satisfy a finite recurrence relation, check out http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/Fibonomials.html. However I have no idea on how to prove this precise formula.

Edit: As Phira mentioned and I forgot to check, it is easily proven using Binet's formula, but a proof that makes no use of irrational numbers would be nice.

$\endgroup$
  • $\begingroup$ Where did you get stuck in your attempt? Please show the work you tried prior to getting stuck. $\endgroup$ – Namaste Jun 19 '18 at 19:55
  • $\begingroup$ It is known that Fibonacci powers for a fixed exponent must satisfy a finite recurrence relation, check out http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/Fibonomials.html. However I have no idea on how to prove this precise formula. $\endgroup$ – Nicolas Malebranche Jun 19 '18 at 20:07
  • $\begingroup$ Remark: the formula generalizes math.stackexchange.com/q/1502690/301936, using that $F_k(x)=\frac{f_k}{1-l_kx-x^2} $ and $L_k(x)=\frac{l_k-2x}{1-l_kx+x^2}$, for $k>0$. $\endgroup$ – Nicolas Malebranche Jun 19 '18 at 20:29
  • $\begingroup$ Just plug in the formula with the golden ratio on the left side. $\endgroup$ – Phira Jun 20 '18 at 7:34
  • $\begingroup$ @Phira Oh thanks, Binet's formula does the job. I should have tried this before. $\endgroup$ – Nicolas Malebranche Jun 20 '18 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.