A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

1, 2, 2, 2, 2, 2, 2, 1, 2, 3, 3, 3, 4, 3, 2, 2, 2, 1, 2, 2, 4, 4, 4, 2, 2, 3, 2, 1, 4, 4, 4, 4, 4, 2, 1, 3, 4, 3, 4, 4, 4, 5, 3, 2, 3, 4, 2, 4, 5, 3, 2, 4, 2, 1, 1, 3, 4, 6, 4, 2, 3, 4, 2, 3, 5, 4, 5, 7, 5, 2, 4, 4, 4, 3, 3, 4, 6, 4, 4, 2, 2, 2, 4, 3, 6, 6, 5, 4, 6, 3, 2, 3, 6, 4, 6, 4, 4, 4, 4, 3, 3

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n >= 0. In other words, each nonnegative integer can be written as the sum of a nonnegative cube and three tetrahedral numbers.

It seems that a(n) = 1 only for n = 0, 7, 17, 27, 34, 53, 54, 110, 118, 163, 207, 263, 270, 309, 362, 443, 1174, 1284.

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

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Example

a(0) = 1 with 0 = 0^3 + C(2,3) + C(2,3) + C(2,3).
a(17) = 1 with 17 = 2^3 + C(3,3) + C(4,3) + C(4,3).
a(27) = 1 with 27 = 3^3 + C(2,3) + C(2,3) + C(2,3).
a(362) = 1 with 362 = 0^3 + C(6,3) + C(8,3) + C(13,3).
a(443) = 1 with 443 = 3^3 + C(5,3) + C(10,3) + C(13,3).
a(1174) = 1 with 1174 = 1^3 + C(9,3) + C(10,3) + C(19,3).
a(1284) = 1 with 1284 = 10^3 + C(7,3) + C(9,3) + C(11,3).

Mathematica

f[n_]:=f[n]=Binomial[n+2, 3];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
tab={}; Do[r=0; Do[If[f[x]>n/3, Goto[cc]]; Do[If[f[y]>(n-f[x])/2, Goto[bb]]; Do[If[f[z]>n-f[x]-f[y], Goto[aa]]; If[CQ[n-f[x]-f[y]-f[z]], r=r+1], {z, y, n-f[x]-f[y]}]; Label[aa], {y, x, (n-f[x])/2}]; Label[bb], {x, 0, n/3}]; Label[cc]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000292, A000578, A000797, A262813, A306460, A306462, A306471, A306477.

Sequence in context: A297033 A342227 A194318 * A297788 A194342 A230410

Adjacent sequences: A306456 A306457 A306458 * A306460 A306461 A306462

Keyword

nonn

Author

Zhi-Wei Sun, Feb 20 2019

Status

approved