A060651 Smallest odd prime p such that Q(sqrt(-p)) has class number 2n+1.

3, 23, 47, 71, 199, 167, 191, 239, 383, 311, 431, 647, 479, 983, 887, 719, 839, 1031, 1487, 1439, 1151, 1847, 1319, 3023, 1511, 1559, 2711, 4463, 2591, 2399, 3863, 2351, 3527, 3719, 3119, 5471, 2999, 4703, 6263, 4391, 3671, 3911, 4079, 5279, 6311, 4679

Offset: 0

Author

Robert G. Wilson v, Apr 17 2001

Keywords

• nonn

Comments

Note that all such primes are congruent to 3 modulo 4.

Conjecture: a(n) = A002146(n) for all n >= 1. That is to say, A002148(n) > A002146(n) for all n >= 1. - Jianing Song, Jul 20 2022

From Jianing Song, Sep 16 2022: (Start)

Note that an imaginary quadratic field has an odd class number if and only if it is of the form Q(sqrt(-1)), Q(sqrt(-2)), or Q(sqrt(-p)) for primes p == 3 (mod 4).

It seems that for most n, the class group of Q(sqrt(-a(n))) is the cyclic group of order 2*n+1. But this is not always true. The smallest prime p such that Q(sqrt(-p)) has class number 243 is p = 29399, and the class group of Q(sqrt(-29399)) is C_3 X C_81 rather than C_243. Also, the smallest prime p such that Q(sqrt(-p)) has class number 637 is p = 149519, and the class group of Q(sqrt(-149519)) is C_7 X C_91 rather than C_637. (End)

Crossrefs

Cf. A002146, A002148.

Programs

Mathematica

<< NumberTheory`NumberTheoryFunctions`
a = Table[0, {101}]; Do[ c = ClassNumber[ -Prime[n] ]; If[ c < 102 && a[ [c] ] == 0, a[ [c] ] = Prime[n] ], {n, 2, 4000} ]; Table[ a[ [n] ], {n, 1, 101} ]
a = Table[0, {101}]; Do[c = NumberFieldClassNumber[Sqrt[-Prime[n]]]; If[c < 102 && a[[c]] == 0, a[[c]] = Prime[n]], {n, 2, 4000}]; Select[ Table[a[[n]], {n, 1, 101}], Mod[#, 4] == 3 &] (* Jean-François Alcover, Jul 20 2022 *)

PARI

a(n) = forprime(p=3, oo, if ((p % 4) == 3, if (qfbclassno(-p) == 2*n+1, return(p)))); \\ Michel Marcus, Jul 20 2022

Formula

a(n) = min(A002146(n), A002148(n)). - Jianing Song, Jul 20 2022

Extensions

Offset corrected by Michel Marcus, Jul 20 2022