OFFSET
0,10
COMMENTS
Conjecture: a(n) > 0 for all n > 2. In other words, each n = 0,1,2,... can be written as x^2 + 2y*(y+2) + z*(z+3)/2 with x,y,z nonnegative integers.
As pointed out by Sun in his 2007 paper in Acta Arith., a result of Jones and Pall implies that every n = 0,1,2,... can be written as x^2 + 2*(2y)^2 + z*(z+1)/2 with x,y,z nonnegative integers.
Let a,c,e be positive integers, and let b,d,f be nonnegative integers with a-b, c-d, e-f all even. Suppose that a|b, c|d and e|f. The author studied in arXiv:1502.03056 when each nonnegative integer can be written as x*(a*x+b)/2 + y*(c*y+d)/2 + z*(e*z+f)/2 with x,y,z nonnegative integers, and conjectured that the answer is positive if (a,b,c,d,e,f) is among the following ten tuples (4,0,2,0,1,3), (4,0,2,0,1,5), (4,0,2,6,1,1), (4,0,2,6,2,0), (4,4,2,0,1,3), (4,8,2,0,1,1), (4,8,2,0,1,3), (4,12,2,0,1,1), (6,0,2,0,1,3), (6,6,2,0,1,3).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
B. W. Jones and G. Pall, Regular and semi-regular positive ternary quadratic forms, Acta Math. 70 (1939), 165-191.
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
EXAMPLE
a(10) = 1 since 10 = 1^2 + 2*2^2 + 1*2/2.
a(11) = 1 since 11 = 0^2 + 2*2^2 + 2*3/2.
a(16) = 1 since 16 = 2^2 + 2*1^2 + 4*5/2.
a(26) = 1 since 26 = 3^2 + 2*1^2 + 5*6/2.
a(31) = 1 since 31 = 1^2 + 2*1^2 + 7*8/2.
a(41) = 1 since 41 = 6^2 + 2*1^2 + 2*3/2.
MATHEMATICA
TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^2-2y^2], r=r+1], {x, 0, Sqrt[n]}, {y, 1, Sqrt[(n-x^2)/2]}]; Print[n, " ", r], {n, 0, 70}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jul 27 2017
STATUS
approved