A254631 Number of ways to write n as x*(x+1)/2 + y*(3*y+2) + z*(3*z-2) with x,y,z nonnegative integers.

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

Offset: 0

Author

Zhi-Wei Sun, Feb 03 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n, and a(n) > 1 for all n > 35.

We have proved that any nonnegative integer n can be written as x*(x+1)/2 + y*(3*y+2) + z*(3*z-2) with x,y,z integers.

Examples

			a(13) = 1 since 13 = 0*1/2 + 1*(3*1+2) + 2*(3*2-2).
a(30) = 1 since 30 = 3*4/2 + 2*(3*2+2) + 2*(3*2-2).
a(35) = 1 since 35 = 1*2/2 + 3*(3*3+2) + 1*(3*1-2).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Crossrefs

Cf. A000217, A000567, A045944, A254573, A254574.

Programs

Mathematica

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