A281826 Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + p(k), where x,y,z are nonnegative integers, k is a positive integer, and p(.) is the partition function given by A000041.

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

Offset: 1

Author

Zhi-Wei Sun, Jan 31 2017

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0.

(ii) Any positive integer n can be written as x^3 + 2*y^3 + 4*z^3 + p(k) with x,y,z nonnegative integers and k a positive integer.

(iii) For each c = 3, 4, any positive integer n can be written as x^3 + 2*y^3 + c*z^3 + A000009(k) with x,y,z nonnegative and k a positive integer.

We have verified the conjecture for n up to 1.3*10^6.

On the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified all the three parts of the conjecture for n up to 10^9. - Zhi-Wei Sun, Feb 06 2017

Examples

			a(1) = 1 since 1 = 0^3 + 2*0^3 + 3*0^3 + p(1).
a(2) = 2 since 2 = 1^3 + 2*0^3 + 3*0^3 + p(1) = 0^3 + 2*0^3 + 3*0^3 + p(2).
a(75) = 3 since 75 = 4^3 + 2*0^3 + 3*0^3 + p(6) = 3^3 + 2*1^3 + 3*2^3 + p(8) = 0^3 + 2*2^3 + 3*1^3 + p(11).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000009, A000041, A000578, A280455.

Programs

Mathematica

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
p[n_]:=p[n]=PartitionsP[n]
Do[r=0;Do[If[p[k]>n,Goto[bb]];Do[If[CQ[n-p[k]-3x^3-2y^3],r=r+1],{x,0,((n-p[k])/3)^(1/3)},{y,0,((n-p[k]-3x^3)/2)^(1/3)}];Continue,{k,1,n}];Label[bb];Print[n," ",r];Continue,{n,1,80}]