A327448 Number of ways the first n cubes can be partitioned into three sets with equal sums.

1, 0, 0, 691, 3416, 0, 233, 1168, 0, 8857, 18157, 0, 2176512, 3628118, 0, 3204865, 8031495, 0, 79514209, 205927212, 0, 5152732369, 13493840291, 0

Offset

23,4

Comments

Note the offset.

References

Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019.

Links

Table of n, a(n) for n=23..46.

Formula

a(n) > 0 => n in { A007494 }. - Alois P. Heinz, Sep 30 2019

Example

The unique smallest solution (for n = 23) is
27 + 216 + 1000 + 2197 + 5832 + 6859 + 9261 =
1 + 64 + 343 + 512 + 1728 + 4096 + 8000 + 10648 =
8 + 125 + 729 + 1331 + 2744 + 3375 + 4913 + 12167.

Maple

s:= proc(n) option remember; `if`(n<2, 0, n^3+s(n-1)) end:
b:= proc(n, x, y) option remember; `if`(n=1, 1, (p-> (l->
      add(`if`(p>l[i], 0, b(n-1, sort(subsop(i=l[i]-p, l))
            [1..2][])), i=1..3))([x, y, s(n)-x-y]))(n^3))
    end:
a:= n-> `if`(irem(1+s(n), 3, 'q')=0, b(n, q-1, q)/2, 0):
seq(a(n), n=23..27);  # Alois P. Heinz, Sep 30 2019

Mathematica

s[n_] := s[n] = If[n < 2, 0, n^3 + s[n - 1]];
b[n_, x_, y_] := b[n, x, y] = If[n == 1, 1, With[{p = n^3}, Sum[If[p > #[[i]], 0, b[n - 1, Sequence @@ Sort[ReplacePart[#, i -> #[[i]] - p]][[1 ;; 2]]]], {i, 1, 3}]]&[{x, y, s[n] - x - y}]];
a[n_] := a[n] = If[q = Quotient[1 + s[n], 3]; Mod[1 + s[n], 3] == 0, b[n, q - 1, q]/2, 0];
Table[Print[n, " ", a[n]]; a[n], {n, 23, 34}] (* Jean-François Alcover, Nov 08 2020, after Alois P. Heinz *)

Crossrefs

Cf. A000537, A000578, A007494, A112972, A275714, A113263, A327449, A327450.

Sequence in context: A127341 A135316 A160500 * A046753 A033563 A231273

Adjacent sequences: A327445 A327446 A327447 * A327449 A327450 A327451

Keyword

nonn,more

Author

N. J. A. Sloane, Sep 19 2019

Extensions

a(32), a(33), a(35) recomputed and a(36)-a(38) added by Alois P. Heinz, Sep 30 2019

a(39)-a(46) from Bert Dobbelaere, May 15 2021

Status

approved