A067069 Numbers n such that for every k the following condition holds: (*) The number of nonnegative solutions of the Diophantine equation x^2 + n*y^2 = 1+n*k^2 equals one half of the number of divisors of 1+n*k^2 if 1+n*k^2 is not a square and one half of 1 + the number of divisors of 1+n*k^2 if 1+n*k^2 is a square.

2, 3, 4, 6, 8, 10, 12, 16, 18, 22, 24, 28, 30, 40, 42, 48, 58, 60, 70, 72, 78, 88, 102, 112, 120, 130, 168, 190, 210, 232, 240, 280, 312, 330, 408, 462, 520, 760, 840, 1320, 1848

Offset

1,1

Comments

The sequence is conjectural. It depends on a proof of conjecture 1: If condition (*) holds for k and n < u(k), then it holds for k+1 and n, where u(k) is some nondecreasing function of k. Computationally this is confirmed for 1 < k < 300 and n < 100, 2 < k < 250 and n < 238, 3 < k < 250 and n < 282, 5 < k < 200 and n < 652, 8 < k < 200 and n < 2088, 9 < k < 200 and n <= 1000000.

No further terms up to 1000000 were found; this suggests conjecture 2: The sequence is finite and 1848 its last term. Perhaps easier to prove is the stronger conjecture 3: For every n > c there is a k < d such that condition (*) is violated for k and n. Computationally this is confirmed for n <=1000000 and c = 1848, d = 11.

Links

Table of n, a(n) for n=1..41.

Prog

(PARI) {h=50; m=5000; v=vector(m, x, 1); for(k=1, h, for(n=1, m, if(v[n]>0, r=0; for(x=0, sqrtint(n*k^2+1), for(y=0, sqrtint((n*k^2+1-x^2)\n), if(x^2+n*y^2==n*k^2+1, r++))); q=n*k^2+1; d=numdiv(q); s=if(issquare(q), d+1, d)/2; if(r!=s, v[n]=0)))); for(n=1, m, if(v[n]>0, print1(n, ", ")))}

Crossrefs

Sequence in context: A309943 A002093 A179971 * A100497 A191283 A341016

Adjacent sequences: A067066 A067067 A067068 * A067070 A067071 A067072

Keyword

nonn

Author

Holger Stephan (stephan(AT)wias-berlin.de), Feb 18 2002

Extensions

Edited and extended by Klaus Brockhaus, Jun 25 2003

Status

approved