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
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