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

1,2

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

Links

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

Example

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).

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}]

Crossrefs

Cf. A000009, A000041, A000578, A280455.

Sequence in context: A062186 A085763 A273264 * A062985 A168092 A210032

Adjacent sequences: A281823 A281824 A281825 * A281827 A281828 A281829

Keyword

nonn

Author

Zhi-Wei Sun, Jan 31 2017

Status

approved