sum of first consecutive powers.

[David Wilson]

Perhaps a better name would be

a(n) = Smallest k >= 1 such that {1^n, 2^n, 3^n, ..., k^n} can be 
parititioned into two sets with equal sum.

It's not very hard to show that a(n) == 0 or 3 (mod 4) for n >= 1.

I verify your results up to a(5) and find a(6) = 31, a(7) = 39.

----- Original Message ----- 
From: "Floor en Lyanne van Lamoen" <fvanlamoen at planet.nl>
To: <seqfan at ext.jussieu.fr>
Sent: Thursday, October 20, 2005 10:42 AM
Subject: sum of first consecutive powers.


> Sequence fans,
>
> I just submitted a sequence (below) that I think is very short. But within
> the limitations of my computer and my abilities, I didn't succeed to find
> a(6). Any shots?
>
> Remarkably there seem to be 3 different smallest solutions for n=5 - I
> didn't reproduce them.
>
> %I A113246
> %S A113246 2,3,7,12,16,24
> %N A113246 The smallest number such that 1^n +- 2^n +- ... +- a(n)^n = 0 
> has
> a solution.
> %e A113246 For n=2 we have 1^2 + 2^2 - 3^2 + 4^2 - 5^2 - 6^2 + 7^2 = 0 as
> smallest solution, so a(2)=7.
> %O A113246 0
> %K A113246 ,more,nonn,
>
> Kind regards,
> Floor van Lamoen.
>
>