A256132 Number of ways to write n as w*(3w+1)/2 + x*(3x-1)/2 + y*(3y-1)/2 + z*(3z-1)/2, where w,x,y,z are nonnegative integers with x <= y <= z.

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

Offset

0,3

Comments

Conjecture: a(n) > 0 for all n. In other words, any nonnegative integer n can be expressed as the sum of three pentagonal numbers and a second pentagonal number.

See also A255350 for a similar conjecture.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..6000

Zhi-Wei Sun, A result similar to Lagrange's theorem, arXiv:1503.03743 [math.NT], 2015.

Example

 a(4) = 1 since 4 = 1*(3*1+1)/2 + 0*(3*0-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1)/2.
a(11) = 1 since 11 = 0*(3*0+1)/2 + 1*(3*1-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1)/2.
a(23) = 1 since 23 = 0*(3*0+1)/2 + 0*(3*0-1)/2 + 1*(3*1-1)/2 + 4*(3*4-1)/2.

Mathematica

GenPen[n_]:=IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1], 6]==1
Do[r=0; Do[If[GenPen[n-x(3x-1)/2-y(3y-1)/2-z(3z-1)/2], r=r+1], {x, 0, (Sqrt[8n+1]+1)/6}, {y, x, (Sqrt[12(n-x(3x-1)/2)+1]+1)/6},
{z, y, (Sqrt[24(n-x(3x-1)/2-y(3y-1)/2)+1]+1)/6}]; Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000326, A005449, A255350.

Sequence in context: A343384 A237615 A343190 * A340057 A303476 A187201

Adjacent sequences: A256129 A256130 A256131 * A256133 A256134 A256135

Keyword

nonn

Author

Zhi-Wei Sun, Mar 15 2015

Status

approved