cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A298672 Number of ordered ways of writing n^3 as a sum of n positive cubes.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 20, 0, 1121, 72828, 872640, 9037710, 118590450, 1743739426, 24407782672, 424735169040, 7802802463460, 135385454550288, 2823521345232834, 59332856029292241, 1238888844244575904, 28893281420537822022, 684650546073054870188, 16342742577592266281996
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 24 2018

Keywords

Examples

			a(6) = 20 because we have [64, 64, 64, 8, 8, 8], [64, 64, 8, 64, 8, 8], [64, 64, 8, 8, 64, 8], [64, 64, 8, 8, 8, 64], [64, 8, 64, 64, 8, 8], [64, 8, 64, 8, 64, 8], [64, 8, 64, 8, 8, 64], [64, 8, 8, 64, 64, 8], [64, 8, 8, 64, 8, 64], [64, 8, 8, 8, 64, 64], [8, 64, 64, 64, 8, 8], [8, 64, 64, 8, 64, 8], [8, 64, 64, 8, 8, 64], [8, 64, 8, 64, 64, 8], [8, 64, 8, 64, 8, 64], [8, 64, 8, 8, 64, 64], [8, 8, 64, 64, 64, 8], [8, 8, 64, 64, 8, 64], [8, 8, 64, 8, 64, 64] and [8, 8, 8, 64, 64, 64].

Links

Crossrefs

Cf. A000578, A030272, A232173, A259792, A290247, A291700, A298329, A298330, A298641, A298671.

Programs

  • Mathematica
    Join[{1}, Table[SeriesCoefficient[Sum[x^k^3, {k, 1, n}]^n, {x, 0, n^3}], {n, 1, 23}]]

Formula

a(n) = [x^(n^3)] (Sum_{k>=1} x^(k^3))^n.

A298671 Number of ordered ways of writing n^3 as a sum of n nonnegative cubes.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 146, 4207, 26329, 257721, 3556495, 42685181, 631230381, 9409600499, 142557084957, 2781352245050, 52598395446786, 950288577530017, 20768368026768594, 448759012546543804, 9652848877533217174, 235179507693424886403, 5756272592837812726164, 140920987987840184113287
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 24 2018

Keywords

Examples

			a(3) = 3 because we have [27, 0, 0], [0, 27, 0] and [0, 0, 27].

Links

Crossrefs

Cf. A000578, A030272, A232173, A259792, A290054, A290247, A291700, A298329, A298330, A298641, A298672.

Programs

  • Mathematica
    Table[SeriesCoefficient[Sum[x^k^3, {k, 0, n}]^n, {x, 0, n^3}], {n, 0, 23}]

Formula

a(n) = [x^(n^3)] (Sum_{k>=0} x^(k^3))^n.

A298641 Number of partitions of n^3 into cubes > 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 8, 6, 45, 100, 377, 1181, 4063, 13225, 45218, 150928, 511970, 1717140, 5777895, 19308880, 64360153, 213446697, 705095144, 2317573307, 7583418322, 24690176885, 80003762726, 257959340058, 827713115396, 2642967441892, 8398644246488
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 24 2018

Keywords

Examples

			a(4) = 2 because we have [64] and [8, 8, 8, 8, 8, 8, 8, 8].

Links

Crossrefs

Cf. A000578, A003108, A030272, A078128, A092362, A259792, A279329, A280130, A290247.

Programs

  • Maple
    g:= proc(n, L) # number of partitions of n into cubes > 1 and <= L
   option remember;
   local t,k;
   t:= 0;
   if n = 0 then return 1 fi;
   if n < 8 then return 0 fi;
   for k from 2 while k^3 <= min(n,L) do
     t:= t + procname(n-k^3, k^3)
   od
end proc:
f:= n -> g(n^3, n^3):
map(f, [$0..50]); # Robert Israel, Jan 24 2018
  • Mathematica
    mx = 30; s = Series[Product[1/(1 - x^(k^3)), {k, 2, mx}], {x, 0, mx^3}]; Table[ CoefficientList[s, x][[1 + n^3]], {n, 0, mx}] (* Robert G. Wilson v, Jan 24 2018 *)

Formula

a(n) = [x^(n^3)] Product_{k>=2} 1/(1 - x^(k^3)).
a(n) = A078128(A000578(n)).
a(n) ~ exp(4*(Gamma(1/3) * Zeta(4/3))^(3/4) * n^(3/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3))^(3/2) / (8 * 3^(5/2) * Pi^2 * n^6). - Vaclav Kotesovec, Jan 31 2018

A347591 Number of compositions (ordered partitions) of n^3 into at most n cubes.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 27, 553, 3192, 87185, 999959, 10684689, 137722770, 2005577212, 27957554982, 492643033682, 8952039793647, 154671244623527, 3207929433418044, 66983196041550714, 1392059664888123313, 32337888832381327369, 763357156272340549200
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 08 2021

Keywords

Links

Crossrefs

Cf. A000578, A023358, A290247, A298672, A307738, A347590, A347592.

A331899 Number of compositions (ordered partitions) of n^3 into distinct cubes.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 7, 1, 1, 127, 1, 1, 127, 769, 10945, 15961, 86641, 86521, 430717, 4140367, 4146751, 93669001, 1538834041, 663998665, 6883029151, 1014140647, 20591858857, 121532206567, 1637261351983, 2981530899847, 5950338797191, 47072230385425
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 31 2020

Keywords

Examples

			a(6) = 7 because we have [216], [125, 64, 27], [125, 27, 64], [64, 125, 27], [64, 27, 125], [27, 125, 64] and [27, 64, 125].

Links

Crossrefs

Cf. A000578, A030272, A259792, A290247, A298641, A298672, A298848, A331845, A331884.

Programs

  • Maple
    b:= proc(n, i, p) option remember;
      `if`((i*(i+1)/2)^2n, 0, b(n-i^3, i-1, p+1))+b(n, i-1, p)))
    end:
a:= n-> b(n^3, n, 0):
seq(a(n), n=0..33);  # Alois P. Heinz, Jan 31 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = If[(i(i+1)/2)^2 < n, 0, If[n == 0, p!, If[i^3 > n, 0, b[n - i^3, i - 1, p + 1]] + b[n, i - 1, p]]];
a[n_] := b[n^3, n, 0];
a /@ Range[0, 33] (* Jean-François Alcover, Nov 26 2020, after Alois P. Heinz *)

Formula

a(n) = A331845(A000578(n)).

A337989 Number of compositions (ordered partitions) of n^n into n-th powers.

Original entry on oeis.org

1, 2, 120, 131204813713122
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 06 2020

Keywords

Comments

The next term is too large to include.

Examples

			a(3) = 120 because 3^3 = 27 and we have [27], [8, 8, 8, 1, 1, 1] (20 permutations), [8, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] (78 permutations), [8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] (20 permutations), [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] and 1 + 20 + 78 + 20 + 1 = 120.

Links

Crossrefs

Cf. A011782, A224366, A290247, A331402.

Programs

  • Mathematica
    Table[SeriesCoefficient[1/(1 - Sum[x^(k^n), {k, 1, n}]), {x, 0, n^n}], {n, 1, 4}]

Formula

a(n) = [x^(n^n)] 1 / (1 - Sum_{k>=1} x^(k^n)).

A346566 Number of compositions (ordered partitions) of n^4 into fourth powers.

Original entry on oeis.org

1, 1, 2, 14474, 131204813713122, 132431858995585724536583901671708647, 20717425607872632053632351670738034346764162305716050849113959210873829416
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 23 2021

Keywords

Comments

The next term is too large to include.

Links

Crossrefs

Cf. A000583, A224366, A259793, A290247, A337989.

Programs

  • Mathematica
    Table[SeriesCoefficient[1/(1 - Sum[x^(k^4), {k, 1, n}]), {x, 0, n^4}], {n, 0, 6}]

Formula

a(n) = [x^(n^4)] 1 / (1 - Sum_{k>=1} x^(k^4)).
Showing 1-7 of 7 results.

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