A363148 a(n) gives the number of equivalence classes of quaternary quadratic forms of discriminant A363147(n) not representing 2.
1, 1, 2, 1, 1, 2, 3, 4, 1, 2, 2, 1, 1, 4, 6, 2, 6, 5, 7, 1, 1, 7, 4, 2, 9, 10, 7, 13, 5, 8, 11, 3, 5, 15, 3, 5, 7, 6, 8, 14, 20, 3, 4, 17, 6, 9, 8, 15, 10, 19, 20, 26, 7, 20, 20, 12, 34, 7, 13, 32, 26, 10, 16, 16, 23, 11, 17, 41, 37, 11, 28, 46, 20, 28, 14, 17
Offset: 1
Keywords
Examples
a(5) = 1 as there is only one equivalence class of quaternary quadratic form of discriminant A363147(5) = 277 not representing 2 (see A307250).
Links
- Andy Huchala, Table of n, a(n) for n = 1..20000
- Andy Huchala, Growth of A363147(n) vs a(n)
- F. Hirzebruch, Modulflächen und Modulkurven zur symmetrischen Hilbertschen Modulgruppe, Annales scientifiques de l’É.N.S. 4e série, tome 11, no 1 (1978), p. 101-165. See page 135.
- Jürg Kramer, On the linear independence of certain theta-series, Mathematische Annalen 281.2 (1988): 219-228. See page 226.
Programs
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Sage
bound = 100 P = Primes() p = 2 for i in range(bound): p = P.next(p) if p % 4 == 1: K1. = NumberField(x^2 - p) K2. = NumberField(x^2 + p) K3.= NumberField(x^2 + 3*p) zeta = K1.zeta_function() h2 = len(K2.class_group()) h3 = len(K3.class_group()) H_plus = int(abs(.49+1/2*zeta(-1)+1/8 * h2 + 1/6*h3)) H = (H_plus+int((p + 19)/24))/2 if H_plus-H>0: print(H_plus-H)
Comments