OFFSET
1,9
COMMENTS
Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. The conjecture holds if a(p) > 0 for every odd prime p. For any n > 0 we have a(3*n) > 0, since 3*n = n + n + n and 3 + 5 + 8 = 4^2.
It seems that a(n) = 1 only for n = 3..8, 10, 11, 19, 53, 89, 127, 178, 257, 461.
See also A343862 for similar conjectures.
Conjecture 1 holds for all n < 2^15. Note a(1823) = 1. - Martin Ehrenstein, May 03 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 version, arXiv:1604.06723 [math.NT], 2016-2017.
EXAMPLE
a(4) = 1 with 4 = 2 + 1 + 1 and 3*2^2*1^2 + 5*1^2*1^2 + 8*1^2*2^2 = 7^2.
a(19) = 1 with 19 = 9 + 9 + 1 and 3*9^2*9^2 + 5*9^2*1^2 + 8*1^2*9^2 = 144^2.
a(53) = 1 with 53 = 23 + 7 + 23 and 3*23^2*7^2 + 5*7^2*23^2 + 8*23^2*23^2 = 1564^2.
a(89) = 1 with 89 = 2 + 58 + 29 and 3*2^2*58^2 + 5*58^2*29^2 + 8*29^2*2^2 = 3770^2.
a(257) = 1 with 257 = 11 + 164 + 82 and 3*11^2*164^2 + 5*164^2*82^2 + 8*82^2*11^2 = 30340^2.
a(461) = 1 with 461 = 186 + 165 + 110 and 3*186^2*165^2 + 5*165^2*110^2 + 8*110^2*186^2 = 88440^2.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[3x^2*y^2+(n-x-y)^2*(5*y^2+8*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, Apr 24 2021
STATUS
approved