OFFSET
1,2
COMMENTS
If instead we count only primitive quadruples (meaning quadruples (h,i,j,k) with gcd(h,i,j,k) = 1) we get A278085(n).
Conjectures from Colin Mallows, Jun 12 2022: (Start)
Given a natural number n, a "quad" for n is a quadruple q = (h,i,j,k) of integers with sum(q) = h+i+j+k = n and sum(q^2) = h^2+i^2+j^2+k^2 = n^2.
A quad q is "primitive" if gcd(h,i,j,k) = 1. Define pq(n) = A278085(n) to be the number of distinct primitive quads for n, and tq(n) (the present sequence) to be the total number of quads for n.
Conjecture 1: (Based on the data for n <= 5000) pq/4 and tq/4 are multiplicative sequences.
Conjecture 2: When n = p^k, p prime and k >= 1:
if p = 2, k = 1 then pq(q)/4 = 1 and tq(n)/4 = 2;
if p = 2, k >= 2 then pq(q)/4 = 0 and tq(n)/4 = 2;
if p = 3, k >= 1 then pq(q)/4 = n and tq(n)/4 = (3*n-1)/2;
if p == 5 (mod 6), k >= 1 then pq(q)/4 = (p+1)*n/p and tq(n)/4 = n + 2*(n-1)/(p-1);
if p == 1 (mod 6), k >= 1 then pq(q)/4 = (p-1)*n/p and tq(n)/4 = n.
(End)
Conjecture: the numbers n for which a(n) = n have a positive asymptotic density.
LINKS
Robert Israel, Table of n, a(n) for n = 1..650
EXAMPLE
Solutions for n = 1: (1,0,0,0) and all permutations thereof.
n=2: (2,0,0,0) and (1,1,1,-1).
n=3: (3,0,0,0) and (2,2,-1,0).
n=4: (4,0,0,0) and (2,2,2,-2). Eight solutions, so a(4) = 8/4 = 2. None are primitive, so A278085(4) = 0.
n=5: (5,0,0,0) and (4,2,-2,1). 4+24 solutions, so a(5) = 28/4 = 7. 24 are primitive, so A278085(5) = 24/4 = 6.
MAPLE
f:= proc(n) local d; add(g3(n-d, n^2 - d^2), d=-n .. n)/4 end proc:
g3:= proc(x, y) option remember; local m, c;
if x^2 > 3*y then return 0 fi;
m:= floor(sqrt(y));
add(g2(x-c, y - c^2), c=- m.. m)
end proc:
g2:= proc(x, y) option remember;
local v;
v:= 2*y - x^2;
if not issqr(v) then 0
elif v = 0 then 1
else 2
fi
end proc:
map(f, [$1..100]); # Robert Israel, Feb 16 2023
MATHEMATICA
f[n_] := Sum[g3[n - d, n^2 - d^2], {d, -n, n}]/4 ;
g3[x_, y_] := g3[x, y] = Module[{m}, If[x^2 > 3*y, 0, m = Floor[Sqrt[y]]; Sum[g2[x - c, y - c^2], {c, -m, m}]]];
g2[x_, y_] := g2[x, y] = Module[{v}, v = 2*y - x^2; Which[!IntegerQ@Sqrt[v], 0, v == 0, 1, True, 2]];
f /@ Range[100] (* Jean-François Alcover, Mar 09 2023, after Robert Israel *)
CROSSREFS
KEYWORD
nonn,look
AUTHOR
N. J. A. Sloane, Jun 19 2022, based on an email from Colin Mallows, Jun 12 2022
STATUS
approved