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 A327448 Number of ways the first n cubes can be partitioned into three sets with equal sums. 3
 1, 0, 0, 691, 3416, 0, 233, 1168, 0, 8857, 18157, 0, 2176512, 3628118, 0, 3204865, 8031495, 0, 79514209, 205927212, 0, 5152732369, 13493840291, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 23,4 COMMENTS Note the offset. REFERENCES Keith F. Lynch, Posting to Math Fun Mailing List, Sep 17 2019. LINKS 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

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Last modified December 5 00:21 EST 2021. Contains 349530 sequences. (Running on oeis4.)