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
The formula is S(m*(2n-1)+n)=0 mod S(n). This means that the sum of m times (2n-1) plus n is equivalent to 0 when divided by the sum of n.
"Mod" stands for modulus, which is the remainder after dividing one number by another. In this formula, it is used to show the congruence between two numbers.
This formula is related to modular arithmetic because it uses the concept of congruence, which is the basis of modular arithmetic. It shows that two numbers are equivalent when divided by another number, or in other words, they have the same remainder.
Proving this formula is significant because it demonstrates the fundamental principles of modular arithmetic and can be applied to solve various mathematical problems. It also shows the relationship between different numbers and how they can be manipulated using modular arithmetic.
This formula can be proven using mathematical induction. This method involves proving that the formula holds for a base case (usually n=1) and then showing that if it holds for any value of n, it also holds for n+1. By repeatedly applying this process, the formula can be proven for all values of n.