

A255350


Number of ways to write n as a*(2a1)+ b*(2b1) + c*(2c+1) + d*(2d+1), where a,b,c,d are nonnegative integers with a <= b and c <= d.


9



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

0,7


COMMENTS

Conjecture: (i) a(n) > 0 for all n. In other words, any nonnegative integer can be expressed as the sum of two hexagonal numbers and two second hexagonal numbers.
(ii) Each nonnegative integer can be written as the sum of two pentagonal numbers and two second pentagonal numbers.
We have verified parts (i) and (ii) of the conjecture for n up to 10^7 and 10^6 respectively.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 20092015.
ZhiWei Sun, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015.


EXAMPLE

a(23) = 1 since 23 = 1*(2*11) + 1*(2*11) + 0*(2*0+1) + 3*(2*3+1).
a(68) = 1 since 68 = 1*(2*11) + 4*(2*41) + 1*(2*1+1) + 4*(2*4+1).


MATHEMATICA

HQ[n_]:=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1], 4]==1
Do[r=0; Do[If[HQ[nx(2x1)y(2y1)z(2z+1)], r=r+1], {x, 0, (Sqrt[4n+1]+1)/4}, {y, x, (Sqrt[8(nx(2x1))+1]+1)/4}, {z, 0, (Sqrt[4(nx(2x1)y(2y1))+1]1)/4}];
Print[n, " ", r]; Continue, {n, 0, 10000}]


CROSSREFS

Cf. A000326, A000384, A005449, A014105.
Sequence in context: A112933 A270650 A088427 * A104482 A333632 A209332
Adjacent sequences: A255347 A255348 A255349 * A255351 A255352 A255353


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 21 2015


STATUS

approved



