OFFSET
1,6
COMMENTS
Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. Conjecture 1 holds if a(p) > 0 for each odd prime p. For any n > 0 we have a(3*n) > 0 since 3*n = n + n + n and 1 + 5 + 10 = 4^2.
See also A340274 for a similar conjecture.
Conjecture 2: There are infinitely many triples (a,b,c) of positive integers such that each n = 3,4,... can be written as x + y + z with x,y,z positive integers and a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 a square.
Such triple candidates include (21,19,9), (23,17,9), (24,16,9), (25,14,10), (29,19,16), (33,27,21), (35,9,5), (37,32,31) etc.
Conjecture 1 holds for all n < 2^15. - Martin Ehrenstein, May 02 2021
LINKS
Martin Ehrenstein, Table of n, a(n) for n = 1..32767 (first 1500 terms from Zhi-Wei Sun)
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].
EXAMPLE
a(4) = 1 with 4 = 2 + 1 + 1 and 2^2*1^2 + 5*1^2*1^2 + 10*1^2*2^2 = 7^2.
a(5) = 1 with 5 = 1 + 3 + 1 and 1^2*3^2 + 5*3^2*1^2 + 10*1^1*1^2 = 8^2.
a(8) = 1 with 8 = 4 + 2 + 2 and 4^2*2^2 + 5*2^2*2^2 + 10*2^2*4^2 = 28^2.
a(9) = 1 with 9 = 3 + 3 + 3 and 3^2*3^2 + 5*3^2*3^2 + 10*3^2*3^2 = 36^2.
a(19) = 2. We have 19 = 4 + 5 + 10 with 4^2*5^2 + 5*5^2*10^2 + 10*10^2*4^2 = 170^2, and 19 = 4 + 13 + 2 with 4^2*13^2 + 5*13^2*2^2 + 10*2^2*4^2 = 82^2.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[x^2*y^2+(n-x-y)^2*(5*y^2+10*x^2)], r=r+1], {x, 1, n-2}, {y, 1, n-1-x}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 02 2021
STATUS
approved