A262815 Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.

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

Offset: 0

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 = 0, 5, 12, 19, 20, 75, 83, 97, 117.

Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 20 2023

See also A262813 and A262816 for similar conjectures.

By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.

Examples

			a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Crossrefs

Cf. A000217, A000578, A005449, A160325, A262813, A262816.

Programs

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0;Do[If[TQ[n-x^3-z(3z+1)/2],r=r+1],{x,0,n^(1/3)},{z,0,(Sqrt[24(n-x^3)+1]-1)/6}];Print[n," ",r];Continue,{n,0,100}]