applications of second order recurrence relation formula
We give two applications of the formula for sequences satisfying second order recurrence relations:

1.
Recall that the Fibonacci sequence^{} satisfies the recurrence relation
$${f}_{n+1}={f}_{n}+{f}_{n1}.$$ Thus, ${f}_{0}=1$, $A=1$, and $B=1$. Therefore, the theorem yields the following formula^{} for the Fibonacci sequence:
$${f}_{n}=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor}\left(\genfrac{}{}{0pt}{}{nk}{k}\right)$$ 
2.
Fix (http://planetmath.org/Fix2) a prime $p$ and define a sequence^{} $s$ by ${s}_{n}=\tau ({p}^{n})$, where $\tau $ denotes the Ramanujan tau function^{}. Recall that $\tau $ satisfies
$$\tau ({p}^{n+1})=\tau (p)\tau ({p}^{n}){p}^{11}\tau ({p}^{n1}).$$ Thus, ${s}_{0}=1$, $A=\tau (p)$, and $B={p}^{11}$. Therefore, the theorem yields
$$\tau ({p}^{n})=\sum _{k=0}^{\lfloor \frac{n}{2}\rfloor}\left(\genfrac{}{}{0pt}{}{nk}{k}\right){({p}^{11})}^{k}{(\tau (p))}^{n2k}.$$ This formula is valid for all primes $p$ and all nonnegative integers $n$.
Title  applications of second order recurrence relation formula 
Canonical name  ApplicationsOfSecondOrderRecurrenceRelationFormula 
Date of creation  20130322 17:51:46 
Last modified on  20130322 17:51:46 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  8 
Author  Wkbj79 (1863) 
Entry type  Application 
Classification  msc 11A25 
Classification  msc 11F11 
Classification  msc 11B39 
Classification  msc 11B37 
Classification  msc 03D20 
Related topic  FibonacciSequence 
Related topic  RamanujanTauFunction 