I think 25, the best equation I have so far is conditional so probably wrong but is F(n) = [F(n-1) +2^n] modulo 18 for n <4, F(n) = [F(n-1) +2^n] modulo 36 for n > 4.
I think 25, the best equation I have so far is conditional so probably wrong but is F(n) = [F(n-1) +2^n] modulo 18 for n <4, F(n) = [F(n-1) +2^n] modulo 36 for n > 4.
First of, you are the first one who has the courage to give it a try. Nice. The solution is a single function for all n. The sequence looks like 2^n variant, but it isn't. A small hint: F(n) = ... +1.
Thanks for the tip, I've got it now- 7. You sum the digits of the previous value and then add 1.