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

%I #25 May 19 2023 11:07:15

%S 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 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 26,10,16,16,23,11,17,41,37,11,28,46,20,28,14,17

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

%C 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 Andy Huchala, <a href="/A363148/b363148.txt">Table of n, a(n) for n = 1..20000</a>

%H Andy Huchala, <a href="/A363148/a363148.pdf">Growth of A363147(n) vs a(n)</a>

%H 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 Jürg Kramer, <a href="http://www.digizeitschriften.de/dms/resolveppn/?PID=GDZPPN00233125X">On the linear independence of certain theta-series</a>, Mathematische Annalen 281.2 (1988): 219-228. See page 226.

%e 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 (Sage)

%o bound = 100

%o P = Primes()

%o p = 2

%o for i in range(bound):

%o p = P.next(p)

%o if p % 4 == 1:

%o K1.<a> = NumberField(x^2 - p)

%o K2.<b> = NumberField(x^2 + p)

%o K3.<c> = NumberField(x^2 + 3*p)

%o zeta = K1.zeta_function()

%o h2 = len(K2.class_group())

%o h3 = len(K3.class_group())

%o H_plus = int(abs(.49+1/2*zeta(-1)+1/8 * h2 + 1/6*h3))

%o H = (H_plus+int((p + 19)/24))/2

%o if H_plus-H>0:

%o print(H_plus-H)

%Y Cf. A307250, A363147.

%K nonn

%O 1,3

%A _Andy Huchala_, May 17 2023