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

0,2

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.

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.

Example

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).

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}]

Crossrefs

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

Sequence in context: A133364 A063420 A347917 * A029385 A185155 A249095

Adjacent sequences: A254628 A254629 A254630 * A254632 A254633 A254634

Keyword

nonn

Author

Zhi-Wei Sun, Feb 03 2015

Status

approved