A307981 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.
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, 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, 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
Offset: 0
Examples
a(19) = 1 with 19 = 0^3 + 2*2^3 + 3*1^3 + 0*1*2/6. a(22) = 1 with 22 = 0^3 + 2*1^3 + 3*0^3 + 4*5*6/6. a(112) = 1 with 112 = 3^3 + 2*0^3 + 3*3^3 + 2*3*4/6. a(158) = 1 with 158 = 3^3 + 2*4^3 + 3*1^3 + 0*1*2/6. a(791) = 1 with 791 = 1^3 + 2*5^3 + 3*5^3 + 9*10*11/6. a(956) = 1 with 956 = 9^3 + 2*0^3 + 3*4^3 + 5*6*7/6. a(6363) = 1 with 6363 = 10^3 + 2*13^3 + 3*0^3 + 17*18*19/6.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Programs
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Mathematica
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];f[w_]:=f[w]=Binomial[w+2,3]; 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]
Comments