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A263849 Let R = Z[(1+sqrt(5))/2] denote the ring of integers in the real quadratic number field of discriminant 5. Then a(n) is the largest integer k such that every totally positive element nu in R of norm m = A031363(n) can be written as a sum of three squares in R in at least k ways.

%I A263849 #47 Apr 18 2025 09:53:16
%S A263849 1,6,12,24,32,24,54,24,24,30,24,48,48,96,24,48,96,48,24,120,108,48,72,
%T A263849 48,120,54,48,48,48,84,72,120,72,78,48,144,72,72,128,192,120,96,48,48,
%U A263849 96,96,216,72,48,120,96,96,48,96,48,120,96,224,72,120,48,288,72,48,72,246,240,120,144
%N A263849 Let R = Z[(1+sqrt(5))/2] denote the ring of integers in the real quadratic number field of discriminant 5. Then a(n) is the largest integer k such that every totally positive element nu in R of norm m = A031363(n) can be written as a sum of three squares in R in at least k ways.
%C A263849 Let R = Z[(1+sqrt{5})/2] denote the ring of integers in the real quadratic number field of discriminant 5. The main result of Maass (1941) is that every totally positive nu in R is a sum of 3 squares x^2+y^2+z^2 with x,y,z in R. The number N_{nu} of such representations is given by the formula in the theorem on page 191. The norms of the totally positive elements nu are rational integers m belonging to A031363, so we can order the terms of the sequence according to the values m = A031363(n). [Comment based on remarks from Gabriele Nebe.]
%C A263849 The terms were computed with the aid of Magma by David Durstoff, Nov 11 2015.
%C A263849 The attached file from David Durstoff gives list of pairs m=A031363(n), a(n), and also the initial terms of Maass's series theta(tau).  David Durstoff says: "I expressed theta(tau) in terms of two variables q1 and q2. The coefficient of q1^k q2^m is a(nu) with k = trace(nu/delta) and m = trace(nu), where delta = (5+sqrt{5})/2 is a generator of the different ideal. I computed the terms for q1^0 to q1^10 and all possible powers of q2."
%C A263849 Note that there are examples of totally positive elements x and y in R which have the same norm, but for which the number of ways of writing x as a sum of three squares in R is different to the number of ways of writing y as a sum of three squares in R.  E.g. there are 78 ways of writing (27 + 7*sqrt(5))/2 as a sum of three squares in R, and there are 192 ways of writing 11 as a sum of three squares in R, yet both elements have norm 121.  The sequence of possible norms for which this can occur is 121, 209, 341, 361, 451, 484, 551, 589, 605, 649, ... - _Robin Visser_, Mar 28 2025
%D A263849 Maass, Hans. Über die Darstellung total positiver Zahlen des Körpers R (sqrt(5)) als Summe von drei Quadraten, Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg. Vol. 14. No. 1, pp. 185-191, 1941.
%H A263849 Robin Visser, <a href="/A263849/b263849.txt">Table of n, a(n) for n = 0..6000</a>
%H A263849 Harvey Cohn, <a href="https://doi.org/10.2307/2002024">A computation of some bi-quadratic class numbers</a>, Math. Tables Aids Comput. 12 (1958), 213-217.  See pages 215-217.
%H A263849 David Durstoff, <a href="/A263849/a263849.txt">Table showing list of pairs m=A031363(n), a(n)  </a>
%e A263849 From _Robin Visser_, Mar 30 2025: (Start)
%e A263849 a(1) = 6, as every totally positive element of norm A031363(1)=1 in R can be written as a sum of three squares in R in exactly 6 ways.  E.g. the element 1 in R has norm 1 and can be written as a sum of three squares in R as:
%e A263849   1 = 1^2 + 0^2 + 0^2 = (-1)^2 + 0^2 + 0^2 = 0^2 + 1^2 + 0^2 = 0^2 + (-1)^2 + 0^2 = 0^2 + 0^2 + 1^2 = 0^2 + 0^2 + (-1)^2.
%e A263849 a(2) = 12, as every totally positive element of norm A031363(2)=4 in R can be written as a sum of three squares in R in exactly 12 ways. E.g. the element 2 in R has norm 4 and can be written as a sum of three squares in R as:
%e A263849   2 = 1^2 + 1^2 + 0^2 = 1^2 + 0^2 + 1^2 = 0^2 + 1^2 + 1^2 = 1^2 + (-1)^2 + 0^2 = 1^2 + 0^2 + (-1)^2 = 0^2 + 1^2 + (-1)^2 = (-1)^2 + 1^2 + 0^2 = (-1)^2 + 0^2 + 1^2 = 0^2 + (-1)^2 + 1^2 = (-1)^2 + (-1)^2 + 0^2 = (-1)^2 + 0^2 + (-1)^2 = 0^2 + (-1)^2 + (-1)^2.
%e A263849 a(3) = 24, as every totally positive element of norm A031363(3)=5 in R can be written as a sum of three squares in R in exactly 24 ways. E.g. the element (5+sqrt(5))/2 in R has norm 5 and can be written as a sum of three squares in R as:
%e A263849   (5+sqrt(5))/2 = w^2 + 1^2 + 0^2 = w^2 + 0^2 + 1^2 = 0^2 + w^2 + 1^2 = 1^2 + w^2 + 0^2 = 0^2 + 1^2 + w^2 = 1^2 + 0^2 + w^2 = (-w)^2 + 1^2 + 0^2 = (-w)^2 + 0^2 + 1^2 = 0^2 + (-w)^2 + 1^2 = 1^2 + (-w)^2 + 0^2 = 0^2 + 1^2 + (-w)^2 = 1^2 + 0^2 + (-w)^2 = w^2 + (-1)^2 + 0^2 = w^2 + 0^2 + (-1)^2 = 0^2 + w^2 + (-1)^2 = (-1)^2 + w^2 + 0^2 = 0^2 + (-1)^2 + w^2 = (-1)^2 + 0^2 + w^2 = (-w)^2 + (-1)^2 + 0^2 = (-w)^2 + 0^2 + (-1)^2 = 0^2 + (-w)^2 + (-1)^2 = (-1)^2 + (-w)^2 + 0^2 = 0^2 + (-1)^2 + (-w)^2 = (-1)^2 + 0^2 + (-w)^2, where w = (1+sqrt(5))/2. (End)
%Y A263849 Cf. A031363 (the norms), A035187 (number of ideals with that norm).
%Y A263849 See A263850 for another version of this sequence.
%Y A263849 Cf. A005875 (sum of 3 squares in Z), A000118 (sum of 4 squares in Z).
%K A263849 nonn
%O A263849 0,2
%A A263849 _N. J. A. Sloane_, Nov 15 2015
%E A263849 More terms from _Robin Visser_, Mar 28 2025