Euler in Babylon

A real recursion

May 23, 2015

For every real number $a \gt 1$ is given the sequence $g_a$ by: $g_{a}(x)=1$ for $x \lt a$ $g_{a}(x)=g_{a}(x-1)+g_a(x-a)$ for $x \ge a$

$G(n)=g_{\sqrt {n}}(n)$ $G(90)=7564511$.

Find $\sum G(p)$ for $p$ prime and $10000000 \lt p \lt 10010000$ Give your answer modulo 1000000007.


gamwe6

Written by gamwe6 who lives and works in San Francisco building useful things. You should follow him on Twitter