Posted by Robo123 on Saturday, December 11, 2010 at 1:06pm.
This is a recursive relationship very similar to the Fibonacci numbers, except that A1=2 and A2=5, whereas F1=1 and F2=1 for Fibonacci numbers.
We start with
An=An-1+An-2
Substitute An-1=An-2+An-3, we get
An=2An-2+An-3
Similarly, and continuing,
An=3An-3+2An-4
An=5An-4+3An-5
An=8An-5+5An-6
An=11An-6+8An-5
....
Replacing the coefficients by Fibonacci numbers, where
F0=0, F1=1, F2=1, F3=2, F4=3...
and continuing the substitution until A2 and A1, we would get:
An=Fn-2*A1+Fn-1*A2
Thus
A4=F2*A1+F3*A2
=1*2+2*5
=12
A8
=F6*2+F7*5
=8*2+13*5
=16+65
=81
Work out A8 manually as a check.