OFFSET
1,13
COMMENTS
Conjecture: a(n) > 0 for all n > 17.
Note that a(4*k) > 0 for all k > 0 since 4*k = k + k + 2*k and k^2 + k^2 + 2*k^2 = (2*k)^2.
See also A230121 for a related conjecture.
The conjecture was confirmed by Chao Haung and Zhi-Wei Sun in 2021. - Zhi-Wei Sun, May 09 2021
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..4000
Chao Huang and Zhi-Wei Sun, On partitions of integers with restrictions involving squares, arXiv:2105.03416 [math.NT], 2021.
Zhi-Wei Sun, Diophantine problems involving triangular numbers and squares, a message to Number Theory List, Oct. 11, 2013.
EXAMPLE
a(9) = 1 since 9 = 1 + 4 + 2*2 with 1^2 + 4^2 + 2*2^2 = 5^2.
a(21) = 1 since 21 = 5 + 8 + 2*4 with 5^2 + 8^2 + 2*4^2 = 11^2.
a(34) = 1 since 34 = 7 + 25 + 2*1 with 7^2 + 25^2 + 2*1^2 = 26^2.
a(56) = 1 since 56 = 14 + 14 + 2*14 with 14^2 + 14^2 + 2*14^2 = 28^2.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]]
a[n_]:=Sum[If[SQ[2i^2+j^2+(n-2i-j)^2], 1, 0], {i, 1, (n-2)/2}, {j, 1, (n-2i)/2}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 29 2013
STATUS
approved