A344073 Smallest number k such that C(-k) is the cyclic group of order n, where C(D) is the class group of the quadratic field with discriminant D; or 0 if no such k exists.

3, 15, 23, 39, 47, 87, 71, 95, 199, 119, 167, 327, 191, 215, 239, 407, 383, 335, 311, 776, 431, 591, 647, 695, 479, 551, 983, 831, 887, 671, 719, 791, 839, 1079, 1031, 959, 1487, 1199, 1439, 1271, 1151, 1959, 1847, 1391, 1319, 2615, 3023, 1751, 1511, 1799

Offset: 1

Author

Jianing Song, May 08 2021

Keywords

• nonn

Comments

Different from A060649.

Conjecture 1: a(n) > 0 for all n;

Conjecture 2: a(n) = o(n^2).

What's the next even term after a(20) = 776 and a(104) = 14024?

Examples

			The smallest k such that c(-k) = C_12 is k = 327, so a(12) = 327.
The smallest k such that c(-k) = C_16 is k = 407, so a(16) = 407.
The smallest k such that c(-k) = C_20 is k = 776, so a(20) = 776.
The smallest k such that c(-k) = C_243 is k = 38231, so a(243) = 38231.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..488

Crossrefs

Cf. A060649, A344072.

Programs

PARI

a(n) = if(n==1, 3, my(d=3); while(!isfundamental(-d) || quadclassunit(-d)[2]!=[n], d++); d)

Formula

For odd n, if a(n) > 0, then a(n) >= A060649(n). The smallest odd n such that the inequality is strict is n = 243.

For even n, if a(n) > 0, A060649(n) > 0 and A344072(n/2) > 0, then a(n) >= min{A060649(n), A344072(n/2)/4}. Assuming Conjecture 2 in A344072, we have a(n) >= A060649(n). The smallest n == 2 (mod 4) such that the inequality is strict is n = 342.