A262813 Number of ordered ways to write n as x^3 + y^2 + z*(z+1)/2 with x >= 0, y >=0 and z > 0.

1, 2, 2, 2, 2, 2, 3, 2, 1, 4, 5, 3, 2, 2, 5, 3, 2, 4, 4, 4, 1, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 5, 3, 4, 1, 4, 9, 6, 4, 4, 3, 3, 3, 3, 7, 8, 4, 3, 3, 3, 3, 5, 7, 5, 5, 4, 4, 4, 4, 4, 3, 4, 3, 8, 6, 4, 8, 3, 4, 5, 8, 7, 5, 5, 5, 3, 2, 8, 8, 6, 4, 7, 8, 2, 5, 7, 4, 6, 2, 5, 7, 10, 6, 5, 7, 3, 5, 1, 6, 5

Offset: 1

Author

Zhi-Wei Sun, Oct 03 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 21, 35, 98, 152, 306.

This has been verified for all n = 1..2*10^7.

Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 18 2023

If z >= 0, a(n) = 1 only for n = 21, 35, 98, 306. - Mauro Fiorentini, Jul 20 2023

In contrast with the conjecture, in 2015 the author refined a result of Euler by proving that any positive integer can be written as the sum of two squares and a positive triangular number.

See also A262815, A262816 and A262941 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0^3 + 0^2 + 1*2/2.
a(2) = 2 since 2 = 0^3 + 1^2 + 1*2/2 = 1^3 + 0^2 + 1*2/2.
a(6) = 2 since 6 = 0^3 + 0^2 + 3*4/2 = 1^3 + 2^2 + 1*2/2.
a(9) = 1 since 9 = 2^3 + 0^2 + 1*2/2.
a(21) = 1 since 21 = 0^3 + 0^2 + 6*7/2.
a(35) = 1 since 35 = 0^3 + 5^2 + 4*5/2.
a(98) = 1 since 98 = 3^3 + 4^2 + 10*11/2.
a(152) = 1 since 152 = 0^3 + 4^2 + 16*17/2.
a(306) = 1 since 306 = 1^3 + 13^2 + 16*17/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.
• Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Crossrefs

Cf. A000217, A000290, A000578, A254885, A262785, A262815, A262816, A262941.

Programs

Mathematica

TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]