[seqfan] Re: A006451 = n such that n*(n+1)/2+1 is square?

[franktaw at netscape.net]

-----Original Message-----
>From: Joerg Arndt <arndt at jjj.de>

>...
>
>n(n+1)/2 + k = t^2
>
>multiply by 8 then we have
>
>(2n+1)^2 + (8k-7) = 2(2t)^2.

This is a simple arithmetic mistake.  The correct equation is:

(2n+1)^2 + (8k-1) = 2(2t)^2.

and substitute 8k-1 for 8k-7 in the remainder of the discussion.

>Substitute X for 2t and Y for 2n+1. We get the Pell-equation:
>
>(*) 2X^2 - Y^2 = 8k-7
>
>(*) can be solved if and only if the squarefree part of 8k-7 doesn't
>contain factors congruent to 3 or 5 modulo 8.

Which, as edited, confirms my conjecture.

It is gratifying to have one's conjectures proved so promptly.

>...

Franklin T. Adams-Watters