# Proof needed

1. Oct 1, 2011

### ramsey2879

Given the following S(0) = S(1) = 1 and S(n) = 6S(n-1) - S(n-2)
Prove (or disprove for m,n integer) S(m*(2n-1) + n) = 0 mod S(n).
Any suggestions would be appreciated. Thanks, Ken

2. Oct 1, 2011

### ramsey2879

I am not sure but I believe I have a proof: If you extend the sequence backward S(-1) = S(2), S(-2) = S(3) and in general S(1-a)= S(a). Therefor, we have S(-1*(2n-1)+n) = S(-n+1) = S(n) and the sequence Mod S(n) from S(1-n) to S(n) is {0,a,b,...1,1,...b,a,0}. Now S(n+1) = 6*0-a =-a so the next 2n-1 terms are {-a,-b,...-1,-1,...-b,-a,0. Then the original sequence Mod S(n) starts over since 6*0 -(-a) = a. QED??