The code was also searching for the next
solution of S(pk) % k = 0 (i.e. % k, rather than
% pk). So I kept it running until I found a
solution to that one too. After a total of about
5 and a half months it found k =
6,361,476,515,268,337. This sequence is
I kept looking for S(pk) % pk = 0 for the whole
time, and found that the next solution must be
greater than pk = 253,814,097,223,614,463"
In short he found a(6) for A007506 and a(16) for
A045345. After a big clap for his amazing results...
Q. Find a(7) for
A007506 and a(17) for