A278083 a(n) is 1/6 of the number of primitive integral quadruples with sum = 2*m and sum of squares = 2*m^2, where m = 2*n-1.

1, 4, 4, 8, 12, 12, 12, 16, 16, 20, 32, 24, 20, 36, 28, 32, 48, 32, 36, 48, 40, 44, 48, 48, 56, 64, 52, 48, 80, 60, 60, 96, 48, 68, 96, 72, 72, 80, 96, 80, 108, 84, 64, 112, 88, 96, 128, 80, 96, 144, 100, 104, 128, 108, 108, 144, 112, 96, 144, 128, 132, 160

Offset

1,2

Comments

Set b(m) = a(n) for m = 2*n-1, and b(m) = 0 for m even.

Conjecture: b(m) is multiplicative: for k >= 1, b(2^k) = 0; b(p^k) = p^(k-1)*b(p) for p an odd prime; b(p) = p+1 for p == 3 (mod 4); b(p) = p-1 for p == 1 (mod 4). It would be nice to have a proof of this.

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, with the zeros, i.e., the sequence (b(n): n >= 1) shown in the comments above, type gc(1:120, 2, 2), where r = 2 and s = 2. To get the 1/6 of these counts with no zeros, type gc(seq(1,59,2), 2, 2)[,3]/6.]

Colin Mallows, R programs for A278081-A278086.

Example

6*a(2) = 24 = 6*b(3) because of (-1,2,2,3) and (0,1,1,4) (12 permutations each). For example, (-1) + 2 + 2 + 3 = 6 = 2*3 and (-1)^2 + 2^2 + 2^2 + 3^2 = 18 = 2*3^2 (with n = 2 and m = 3 = 2*n - 1).

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[{m = 2n - 1, s}, s = 2m^2; Sum[q[2m - i - j, s - i^2 - j^2, GCD[i, j]] , {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/6];
Table[an = a[n]; Print[n, " ", an]; an, {n, 1, 100}] (* Jean-François Alcover, Sep 20 2020, after Andrew Howroyd *)

Prog

(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(m=2*n-1, s=2*m^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(2*m-i-j, s-i^2-j^2, gcd(i, j)) ))/6} \\ Andrew Howroyd, Aug 02 2018

Crossrefs

Cf. A046897, A278081, A278082, A278084, A278085, A278086.

Sequence in context: A309456 A004024 A292276 * A086663 A003829 A194697

Adjacent sequences: A278080 A278081 A278082 * A278084 A278085 A278086

Keyword

nonn

Author

Colin Mallows, Nov 14 2016

Extensions

Terms a(51) and beyond from Andrew Howroyd, Aug 02 2018

Name and example section edited by Petros Hadjicostas, Apr 21 2020

Status

approved