A254573 Number of ways to write n = x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z nonnegative integers.

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

Offset: 0

Author

Zhi-Wei Sun, Feb 01 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 4, 6, 10, 11, 23.

This has been verified for all n = 0..10^7. We have proved that every nonnegative integer can be written as x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z integers.

Examples

			a(10) = 1 since 10 = 1*2 + 2*(3*2+1)/2 + 1*(3*1-1)/2.
a(11) = 1 since 11 = 2*3 + 0*(3*0+1)/2 + 2*(3*2-1)/2.
a(23) = 1 since 23 = 4*5 + 1*(3*1+1)/2 + 1*(3*1-1)/2.
a(34) = 2 since 34 = 3*4 + 0*(3*0+1)/2 + 4*(3*4-1)/2 = 4*5 + 1*(3*1+1)/2 + 3*(3*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. A000326, A002378, A005449, A254574.

Programs

Mathematica

sQ[n_]:=IntegerQ[Sqrt[4n+1]]
Do[r=0;Do[If[sQ[n-y(3y+1)/2-z(3z-1)/2],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[24(n-y(3y+1)/2)+1]+1)/6}];
Print[n," ",r];Continue,{n,0,70}]