[seqfan] Re: When n^2 divides sum

[Joshua Zucker]

On Tue, Apr 14, 2009 at 1:42 PM, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
> %S A159555 1,6,22
> %N A159555 Those n where n^2 divides A159553(n), where A159553(n) = sum{k=0 to n} binomial(n,k) * GCD(n,k).

I get 1,6,22, 72, 114, 148, 164, 260, 261, 780
and no more terms up to 1000.

> %S A159458 1,2,3,11,33
> %N A159458 Those n where n^2 divides A159068(n), where A159068(n) = sum{k=1 to n} binomial(n,k) * GCD(n,k).

I get 1,2,3,11,33, 309, 665
and no more terms up to 1000.

> Is there a way to determine the values of these sequences besides trial and error division?

Probably, but I don't know it either!

--Joshua Zucker