A256855 Number of ordered ways to write n as x*(3*x-1)/2 + y*(3*y+1)/2 + z*(3*z+1), where x and y are nonnegative integers and z is an integer.

1, 1, 2, 2, 2, 3, 1, 4, 1, 3, 2, 3, 4, 1, 4, 3, 5, 4, 3, 3, 2, 3, 5, 1, 5, 2, 7, 3, 3, 6, 4, 7, 3, 4, 2, 3, 4, 7, 4, 5, 4, 6, 5, 3, 5, 6, 4, 4, 3, 4, 5, 8, 8, 4, 5, 6, 5, 5, 4, 7, 5, 6, 7, 3, 7, 4, 7, 5, 6, 2, 5, 8, 10, 3, 4, 6, 6, 7, 4, 9, 5, 10, 10, 5, 4, 6, 6, 8, 4, 8, 3, 6, 11, 5, 8, 4, 10, 4, 2, 8, 5

Offset

0,3

Comments

Conjecture: a(n) > 0 for all n. Also, any nonnegative integer can be written as x*(3*x-1)/2 + y*(3*y-1) + z*(3*z+1)/2 with x,y nonnegative integers and z an integer, and each n = 0,1,2,... can be expressed as x*(3*x+1)/2 + 3*y*(3*y+1)/2 + z*(3*z+1)/2 with x,y nonnegative integers and z an integer.

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(8) = 1 since 8 = 1*(3*1-1)/2 + 2*(3*2+1)/2 + 0*(3*0+1).
a(23) = 1 since 23 = 3*(3*3-1)/2 + 2*(3*2+1)/2 + 1*(3*1+1).

Mathematica

PQ[n_]:=IntegerQ[Sqrt[12n+1]]
Do[r=0; Do[If[PQ[n-x(3x-1)/2-y(3y+1)/2], r=r+1], {x, 0, (Sqrt[24n+1]+1)/6}, {y, 0, (Sqrt[24(n-x(3x-1)/2)+1]-1)/6}];
Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000326, A001318, A005449, A253187, A254573, A254574.

Sequence in context: A233567 A141059 A135151 * A273943 A256071 A248808

Adjacent sequences: A256852 A256853 A256854 * A256856 A256857 A256858

Keyword

nonn

Author

Zhi-Wei Sun, Apr 11 2015

Status

approved