cp's OEIS Frontend

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.

%I A281826 #16 Jan 11 2018 04:30:35
%S A281826 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,
%T A281826 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,
%U A281826 11,7,5,7,6,5,3,6,4,4,4,9
%N 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.
%C A281826 Conjecture: (i) a(n) > 0 for all n > 0.
%C A281826 (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.
%C A281826 (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.
%C A281826 We have verified the conjecture for n up to 1.3*10^6.
%C A281826 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
%H A281826 Zhi-Wei Sun, <a href="/A281826/b281826.txt">Table of n, a(n) for n = 1..10000</a>
%e A281826 a(1) = 1 since 1 = 0^3 + 2*0^3 + 3*0^3 + p(1).
%e A281826 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).
%e A281826 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).
%t A281826 CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
%t A281826 p[n_]:=p[n]=PartitionsP[n]
%t A281826 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}]
%Y A281826 Cf. A000009, A000041, A000578, A280455.
%K A281826 nonn
%O A281826 1,2
%A A281826 _Zhi-Wei Sun_, Jan 31 2017