The discussion centers on proving the equation S(m*(2n-1) + n) = 0 mod S(n) using the recurrence relation S(n) = 6S(n-1) - S(n-2) with initial conditions S(0) = S(1) = 1. A proposed proof involves extending the sequence backward to establish that S(-1*(2n-1) + n) equals S(n), leading to a cyclical pattern in the sequence modulo S(n). The conclusion suggests that the sequence resets, confirming the original statement.
PREREQUISITESMathematicians, students studying number theory, and anyone interested in advanced sequence analysis and modular arithmetic proofs.
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??ramsey2879 said: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