

A256132


Number of ways to write n as w*(3w+1)/2 + x*(3x1)/2 + y*(3y1)/2 + z*(3z1)/2, where w,x,y,z are nonnegative integers with x <= y <= z.


2



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
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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

ZhiWei Sun, Table of n, a(n) for n = 0..6000
ZhiWei 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*01)/2 + 1*(3*11)/2 + 1*(3*11)/2.
a(11) = 1 since 11 = 0*(3*0+1)/2 + 1*(3*11)/2 + 2*(3*21)/2 + 2*(3*21)/2.
a(23) = 1 since 23 = 0*(3*0+1)/2 + 0*(3*01)/2 + 1*(3*11)/2 + 4*(3*41)/2.


MATHEMATICA

GenPen[n_]:=IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1], 6]==1
Do[r=0; Do[If[GenPen[nx(3x1)/2y(3y1)/2z(3z1)/2], r=r+1], {x, 0, (Sqrt[8n+1]+1)/6}, {y, x, (Sqrt[12(nx(3x1)/2)+1]+1)/6},
{z, y, (Sqrt[24(nx(3x1)/2y(3y1)/2)+1]+1)/6}]; Print[n, " ", r]; Continue, {n, 0, 100}]


CROSSREFS

Cf. A000326, A005449, A255350.
Sequence in context: A305298 A298824 A237615 * A303476 A187201 A262403
Adjacent sequences: A256129 A256130 A256131 * A256133 A256134 A256135


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 15 2015


STATUS

approved



