%I #25 Aug 28 2023 08:20:30
%S 20,56,104,164,296,356,404,584,1172,776,1076,1316,1256,1364,1844,1784,
%T 2456,2504,4916,2756,3176,3416,3764,4424,4436,5924,6296,4616,5144,
%U 5444,10484,6536,9236,7124,7796,7556,12776,9176,8564,10856,11156,10436,11864,12536,14804,13604,13844,16376,15896,13796
%N Smallest even k such that h(k) = 2n, where h(D) is the class number of the quadratic field with discriminant D; or 0 if no such k exists.
%C In other words, a(n) is the smallest even k such that Q(sqrt(k/4)) has class number n; or 0 if no such k exists.
%C Conjecture 1: a(n) > 0 for all n.
%C Conjecture 2: If a(n) > 0 and A060649(2n) > 0, then we have a(n) > A060649(2n). This would imply that all terms in A225060 are odd.
%C Conjecture 3: There exists a positive constant c such that a(n) < c*A060649(2n) for all n.
%C It seems that the ratio a(n)/A060649(2n) reaches its minimum at n = 3. Among the first 250 terms, the maximum of a(n)/A060649(2n) is ~5.3116, which is attained at n = 227.
%H Jianing Song, <a href="/A344072/b344072.txt">Table of n, a(n) for n = 1..250</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ClassNumber.html">Class Number</a>.
%e The smallest even k such that h(k) = 2 is k = 20, so a(1) = 20.
%e The smallest even k such that h(k) = 4 is k = 56, so a(2) = 56.
%e The smallest even k such that h(k) = 12 is k = 356, so a(6) = 356.
%o (PARI) a(n) = my(d=4); while(!isfundamental(d)  qfbclassno(d)!=2*n, d+=4); d
%Y Cf. A060649, A225060.
%K nonn
%O 1,1
%A _Jianing Song_, May 08 2021
