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View Full Version : All Square Numbers Follow a Recursive Series?


IntegrateRSC
Jun19-11, 09:54 PM
Sorry if this is a well known thing, but I've noticed this and decided to see how well known it is, also if there is a way to prove it other than the recursive series.

an=(an-1-an-2+2)+an-1

This recursive series will in fact generate every square number. Take in example:

a0=0
a1=1

So if you use the recursive series above:
a2=(a1-a0+2)+a1
a2=(1-0+2)+1

Any comments? I'm praying I used the 'sub' tags right

robert2734
Jun19-11, 10:13 PM
let's call sqrt(an-1) is x. Then we want to prove that 2x2=(x+1)2+(x-1)2-2. Simple algebra.

Antiphon
Jun19-11, 10:14 PM
Look up the Z transform and it's inverse. You can form recursive solutions corresponding to polynomials of arbitrary order. These sequences correspond to the natural responses of linear discrete-time systems.