Search: id:A067069 Results 1-1 of 1 results found. %I A067069 %S A067069 2,3,4,6,8,10,12,16,18,22,24,28,30,40,42,48,58,60,70,72,78,88,102,112, %T A067069 120,130,168,190,210,232,240,280,312,330,408,462,520,760,840,1320,1848 %N 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. %C A067069 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. %C A067069 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. %o A067069 (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,",")))} %Y A067069 Sequence in context: A139542 A093717 A002093 this_sequence A100497 A088881 A020697 %Y A067069 Adjacent sequences: A067066 A067067 A067068 this_sequence A067070 A067071 A067072 %K A067069 nonn %O A067069 1,1 %A A067069 Holger Stephan (stephan(AT)wias-berlin.de), Feb 18 2002 %E A067069 Edited and extended by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jun 25 2003 Search completed in 0.001 seconds