A278085 1/4 of the number of primitive integral quadruples with sum = 3*n and sum of squares = 3*n^2.

1, 1, 3, 0, 6, 3, 6, 0, 9, 6, 12, 0, 12, 6, 18, 0, 18, 9, 18, 0, 18, 12, 24, 0, 30, 12, 27, 0, 30, 18, 30, 0, 36, 18, 36, 0, 36, 18, 36, 0, 42, 18, 42, 0, 54, 24, 48, 0, 42, 30, 54, 0, 54, 27, 72, 0, 54, 30, 60, 0, 60, 30, 54, 0, 72, 36, 66, 0, 72, 36, 72, 0, 72, 36, 90, 0, 72, 36, 78, 0, 81, 42, 84, 0, 108, 42, 90, 0, 90, 54, 72, 0, 90, 48, 108, 0, 96, 42, 108, 0

Offset: 1

Author

Colin Mallows, Nov 14 2016

Keywords

• nonn

Comments

Conjecture: a(n) is multiplicative, with a(2) = 1, a(2^k) = 0 for k>=2, and for k >= 1 and p an odd prime, a(p^k) = p^(k-1)*a(p), with a(p) = p+1 for p == 5 (mod 6), a(p) = p-1 for p=1 (mod 6), and a(3) = 3. It would be nice to have a proof of this. [See A354766 for additional conjectures. - N. J. A. Sloane, Jun 19 2022]

This is also 1/4 of the number of primitive integral quadruples with sum = n and sum of squares = n^2. See A354766, A354777, A354778 for the total number of solutions. - N. J. A. Sloane, Jun 27 2022

Examples

			For the case r = s = 3, we have 4*a(3) = 12 because of (1,1,3,4) (12 permutations). Indeed, 1 + 1 + 3 + 4 = 9 = 3*3 and 1^2 + 1^2 + 3^2 + 4^2 = 27 = 3*3^2.
For the case r = s = 1, we have again 4*a(3) = 12 because of (3,3,3,3) - (1,1,3,4) = (2,2,0,-1) (12 permutations). Indeed, 2 + 2 + 0 + (-1) = 3 = 1*3 and 2^2 + 2^2 + 0^2 + (-1)^2 = 9 = 1*3^2.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..500
• Petros Hadjicostas, Slight modification of Mallows' R program. [To get the total counts for n = 1 to 120, type gc(1:120, 3, 3), where r = 3 and s = 3. To get the 1/4 of these counts, type gc(1:120, 3, 3)[,3]/4. As stated in the comments, we get the same sequence with r = 1 and s = 1, i.e., we may type gc(1:120, 1, 1)[,3]/4.]
• Colin Mallows, R programs for A278081-A278086.

Crossrefs

Cf. A046897, A278081, A278082, A278083, A278084, A278086, A354766, A353589, A354777, A354778.

Programs

Mathematica

sqrtint = Floor[Sqrt[#]]&;
q[r_, s_, g_] := Module[{d = 2 s - r^2, h}, If[d <= 0, d == 0 && Mod[r, 2] == 0 && GCD[g, r/2] == 1, h = Sqrt[d]; If[IntegerQ[h] && Mod[r+h, 2] == 0 && GCD[g, GCD[(r+h)/2, (r-h)/2]]==1, 2, 0]]] /. {True -> 1, False -> 0};
a[n_] := Module[{s}, s = 3 n^2; Sum[q[3 n - i - j, s - i^2 - j^2, GCD[i, j]] , {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/4];
Table[an = a[n]; Print[n, " ", an]; an, {n, 1, 100}] (* Jean-François Alcover, Sep 20 2020, after Andrew Howroyd *)

PARI

q(r, s, g)={my(d=2*s - r^2); if(d<=0, d==0 && r%2==0 && gcd(g, r/2)==1, my(h); if(issquare(d, &h) && (r+h)%2==0 && gcd(g, gcd((r+h)/2, (r-h)/2))==1, 2, 0))}
a(n)={my(s=3*n^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(3*n-i-j, s-i^2-j^2, gcd(i,j)) ))/4} \\ Andrew Howroyd, Aug 02 2018

Extensions

Example edited by Petros Hadjicostas, Apr 21 2020