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

Joerg Arndt arndt at jjj.de
Sun Oct 11 03:11:02 CEST 2009


The following message form Matthijs Coster
did not show up on on the list:

Hallo Joerg,

> Quick & dirty:
> ? for(k=0,20,q=0;for(n=1,10000,t=n*(n+1)/2+k;if(issquare(t),q=1;break()));
> if(q, print1(k,", ")))

Notice:

n(n+1)/2 + k = t^2

multiply by 8 then we have

(2n+1)^2 + (8k-7) = 2(2t)^2.

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.
Therefore the quick and dirty program can be improved by:

for (k = 1; k < K_0; k++)
{
 n = 8*k-1;
 factor(n,factors);
 for (i = 0; i < factors.length; i++)
   if (((factors[i].a % 8 == 3) || (factors[i].a % 8 == 5)) &&
(factors[i].b % 2 == 1))
     result[k] = "no solution";
   else
     result[k] = "solution";
   fprintf("%d %s\n", k, result[k]);
 }

Greetings,

Matthijs Coster

[...]





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