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A363148 a(n) gives the number of equivalence classes of quaternary quadratic forms of discriminant A363147(n) not representing 2.

%I A363148 #27 Aug 11 2025 08:16:26
%S A363148 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,
%T A363148 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,
%U A363148 26,10,16,16,23,11,17,41,37,11,28,46,20,28,14,17
%N A363148 a(n) gives the number of equivalence classes of quaternary quadratic forms of discriminant A363147(n) not representing 2.
%C A363148 Conjecture: a(n) ~ c * A363147(n) ^ d where d is a constant which is roughly 1.51 and c is one of four constants, depending on the value of A363147(n) mod 24. See plots in files.
%H A363148 Andy Huchala, <a href="/A363148/b363148.txt">Table of n, a(n) for n = 1..20000</a>
%H A363148 Andy Huchala, <a href="/A363148/a363148.pdf">Growth of A363147(n) vs a(n)</a>
%H A363148 F. Hirzebruch, <a href="http://www.numdam.org/item/10.24033/asens.1342.pdf">Modulflächen und Modulkurven zur symmetrischen Hilbertschen Modulgruppe</a>, Annales scientifiques de l’É.N.S. 4e série, tome 11, no 1 (1978), p. 101-165. See page 135.
%H A363148 Jürg Kramer, <a href="https://gdz.sub.uni-goettingen.de/id/PPN235181684_0281">On the linear independence of certain theta-series</a>, Mathematische Annalen 281.2 (1988): 219-228. See page 226.
%e A363148 a(5) = 1 as there is only one equivalence class of quaternary quadratic form of discriminant A363147(5) = 277 not representing 2 (see A307250).
%o A363148 (Sage)
%o A363148 bound = 100
%o A363148 P = Primes()
%o A363148 p = 2
%o A363148 for i in range(bound):
%o A363148     p = P.next(p)
%o A363148     if p % 4 == 1:
%o A363148         K1.<a> = NumberField(x^2 - p)
%o A363148         K2.<b> = NumberField(x^2 + p)
%o A363148         K3.<c> = NumberField(x^2 + 3*p)
%o A363148         zeta = K1.zeta_function()
%o A363148         h2 = len(K2.class_group())
%o A363148         h3 = len(K3.class_group())
%o A363148         H_plus = int(abs(.49+1/2*zeta(-1)+1/8 * h2 + 1/6*h3))
%o A363148         H = (H_plus+int((p + 19)/24))/2
%o A363148         if H_plus-H>0:
%o A363148             print(H_plus-H)
%Y A363148 Cf. A307250, A363147.
%K A363148 nonn
%O A363148 1,3
%A A363148 _Andy Huchala_, May 17 2023