A278082 1/12 of the number of primitive quadruples with sum = n and sum of squares = 3*n^2.

1, 1, 2, 0, 4, 2, 8, 0, 6, 4, 11, 0, 14, 8, 8, 0, 18, 6, 20, 0, 16, 11, 22, 0, 20, 14, 18, 0, 30, 8, 30, 0, 22, 18, 32, 0, 36, 20, 28, 0, 42, 16, 44, 0, 24, 22, 46, 0, 56, 20, 36, 0, 52, 18, 44, 0, 40, 30, 58, 0, 62, 30, 48, 0, 56, 22, 66, 0, 44, 32, 70, 0, 74, 36, 40, 0, 88, 28, 80, 0, 54, 42, 84, 0, 72, 44, 60, 0, 88, 24, 112, 0, 60, 46, 80, 0, 96, 56, 66, 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 (k >= 2); a(p^k) = p^(k-1)*a(p); a(p) = p + 1 for p == (2, 6, 7, 8, 10)(mod 11), a(p) = p - 1 for p == (1, 3, 4, 5, 9)(mod 11); and p(11) = 11. It would be nice to have a proof of this.

This sequence applies also to the case sum = 3*n and ssq = 5*n^2. - Colin Mallows, Nov 30 2016 [Edited by Petros Hadjicostas, Apr 20 2020]

Examples

			For the case r = 1 and r = 3, we have 12*a(3) = 24 because of (-3,1,1,4) and (-1,-1,0,5) (12 permutations each). For example, (-3) + 1 + 1 + 4 = 3 = 1*3 and (-3)^2 + 1^2 + 1^2 + 4^2 = 27 = 3*3^2.
For the case r = 3 and m = 5, we again have 12*a(3) = 24 because of (3,3,3,3) - (-3,1,1,4) = (6,2,2,-1) and (3,3,3,3) - (-1,-1,0,5) = (4,4,3,-2) (12 permutations each). For example, 6 + 2 + 2 + (-1) = 9 = 3*3 and 6^2 + 2^2 + 2^2 + (-1)^2  = 45 = 5*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, 1, 3), where r = 1 and s = 3. To get the 1/12 of these counts, type gc(1:120, 1, 3)[,3]/12. As stated in the comments, we get the same sequence with r = 3 and s = 5, i.e., we may type gc(1:120, 3, 5)[,3]/12.]
• Colin Mallows, R programs for A278081-A278086.

Crossrefs

Cf. A046897, A278081, A278083, A278084, A278085, A278086.

Programs

Mathematica

sqrtint = Floor[Sqrt[#]]&;
q[r_, s_, g_] := Module[{d = 2s - 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 = 3n^2}, Sum[q[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]}]/12];
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(n-i-j, s-i^2-j^2, gcd(i,j)) ))/12} \\ Andrew Howroyd, Aug 02 2018

Extensions

Example section edited by Petros Hadjicostas, Apr 21 2020