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Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + w*(w+1)*(w+2)/6, where x,y,z,w are nonnegative integers.
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%I #14 May 10 2019 11:51:12

%S 1,2,2,3,4,3,3,3,2,2,3,3,3,3,3,2,2,3,2,1,5,4,1,4,4,4,4,5,6,3,5,5,2,4,

%T 4,3,5,5,3,3,4,3,3,2,5,3,3,5,2,2,3,3,5,2,4,4,3,3,5,6,3,5,6,3,4,4,5,7,

%U 5,4,2,3,2,3,2,4,3,3,3,3,4,5,6,8,7,7,6,7,8,6,7,4,5,4,4,2,2,4,4,5,4

%N Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + w*(w+1)*(w+2)/6, where x,y,z,w are nonnegative integers.

%C Conjecture: a(n) > 0 for every nonnegative integer n. In other words, we have {x^3 + 2*y^3 + 3*z^3 + w*(w+1)*(w+2)/6: x,y,z,w = 0,1,2,...} = {0,1,2,...}.

%C We have verified a(n) > 0 for all n = 0..2*10^6.

%H Zhi-Wei Sun, <a href="/A307981/b307981.txt">Table of n, a(n) for n = 0..10000</a>

%e a(19) = 1 with 19 = 0^3 + 2*2^3 + 3*1^3 + 0*1*2/6.

%e a(22) = 1 with 22 = 0^3 + 2*1^3 + 3*0^3 + 4*5*6/6.

%e a(112) = 1 with 112 = 3^3 + 2*0^3 + 3*3^3 + 2*3*4/6.

%e a(158) = 1 with 158 = 3^3 + 2*4^3 + 3*1^3 + 0*1*2/6.

%e a(791) = 1 with 791 = 1^3 + 2*5^3 + 3*5^3 + 9*10*11/6.

%e a(956) = 1 with 956 = 9^3 + 2*0^3 + 3*4^3 + 5*6*7/6.

%e a(6363) = 1 with 6363 = 10^3 + 2*13^3 + 3*0^3 + 17*18*19/6.

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];f[w_]:=f[w]=Binomial[w+2,3];

%t tab={};Do[r=0;w=0;Label[bb];If[f[w]>n,Goto[aa]];Do[If[CQ[n-f[w]-2y^3-3z^3],r=r+1],{y,0,((n-f[w])/2)^(1/3)},{z,0,((n-f[w]-2y^3)/3)^(1/3)}];w=w+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,0,100}];Print[tab]

%Y Cf. A000292, A000578, A262813, A306460, A306477, A306790.

%K nonn,look

%O 0,2

%A _Zhi-Wei Sun_, May 08 2019