Define a sequence of integers s$0,$ s$1,$ s$2,...$ by the initial conditions s$0=0,$ s$1=1,$ and the recurrence

relation

sn$+1=$ sn $+3$sn$1.$

So$,$ the sequence s$0,$ s$1,$ s$2,...$ begins

$0,1,1,4,7,19,40,...$

Prove that for every n $>=0$ we have

sn $=(1/13)(\backslash$alpha $^$n $\backslash$beta $^$n$)$

where

$\backslash$alpha $=(1+13)/2$

and $\backslash$beta $=(113)/2$