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 1-10 of 12 results. Next

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.

A298934 Number of partitions of n^2 into distinct cubes.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 3, 3, 1, 0, 3, 0, 2, 4, 0, 0, 1, 0, 0, 2, 3, 1, 1, 0, 6, 3, 6, 1, 6, 0, 3, 9, 0, 6, 6, 7, 0, 10, 3, 3, 6, 0, 8, 6, 13, 2, 10, 9, 10, 19, 2, 14, 21, 7, 2, 25
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(15) = 2 because we have [216, 8, 1] and [125, 64, 27, 8, 1].

Links

Crossrefs

Cf. A000290, A000578, A030272, A030273, A218495, A259792, A279329, A298672, A298848, A298935.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1,
      `if`(n>i^2*(i+1)^2/4, 0, b(n, i-1)+
      `if`(i^3>n, 0, b(n-i^3, i-1))))
    end:
a:= n-> b(n^2, n):
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 29 2018
  • Mathematica
    Table[SeriesCoefficient[Product[1 + x^k^3, {k, 1, Floor[n^(2/3) + 1]}], {x, 0, n^2}], {n, 0, 84}]

Formula

a(n) = [x^(n^2)] Product_{k>=1} (1 + x^(k^3)).
a(n) = A279329(A000290(n)).

A307643 Number of partitions of n^3 into exactly n positive cubes.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 1, 0, 2, 6, 14, 23, 51, 108, 228, 511, 1158, 2500, 5603, 12304, 26969, 59222, 130115, 285370, 624965, 1368603, 2987117, 6517822, 14187920, 30823278, 66834822, 144671698, 312551894, 673913968, 1450292087, 3114720013, 6676277754, 14281662079
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 19 2019

Keywords

Examples

			9^3 =
1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 =
1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 =
1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 =
1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 =
1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 =
2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 7^3,
so a(9) = 6.

Links

Crossrefs

Cf. A000578, A030272, A259792, A298671, A298672, A307644, A319435, A320841.

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n=0, `if`(t=0, 1, 0),
      `if`(i<1 or t<1, 0, b(n, i-1, t)+
      `if`(i^3>n, 0, b(n-i^3, i, t-1))))
    end:
a:= n-> b(n^3, n$2):
seq(a(n), n=0..25);  # Alois P. Heinz, Oct 12 2019
  • Mathematica
    b[n_, i_, t_] := b[n, i, t] = If[n == 0, If[t == 0, 1, 0], If[i < 1 || t < 1, 0, b[n, i - 1, t] + If[i^3 > n, 0, b[n - i^3, i, t - 1]]]];
a[n_] := b[n^3, n, n];
a /@ Range[0, 25] (* Jean-François Alcover, Nov 07 2020, after Alois P. Heinz *)

Formula

a(n) = A320841(n^3,n).

Extensions

More terms from Vaclav Kotesovec, Apr 20 2019

A299195 Number of ordered ways of writing n^4 as a sum of n fourth powers of positive integers.

Original entry on oeis.org

1, 1, 0, 0, 0, 30, 6, 0, 0, 0, 360, 157080, 0, 12586860, 0, 714233520, 579379361, 48062263014, 46026944529624, 759085890469938, 170947379002578290, 3331302954541376850, 479526242126281889924, 11322897238957194004884, 1341983461418984670506352, 31585668052999315295625900
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 04 2018

Keywords

Examples

			a(6) = 6 because we have [256, 256, 256, 256, 256, 16], [256, 256, 256, 256, 16, 256], [256, 256, 256, 16, 256, 256], [256, 256, 16, 256, 256, 256], [256, 16, 256, 256, 256, 256] and [16, 256, 256, 256, 256, 256].

Links

Crossrefs

Cf. A000583, A046042, A259793, A298330, A298672, A298859, A299169.

Programs

  • Mathematica
    a[0] = 1; a[n_] := Coefficient[Sum[x^k^4, {k, n-1}]^n // Expand, x, n^4]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 25}] (* Jean-François Alcover, Feb 05 2018 *)

Formula

a(n) = [x^(n^4)] (Sum_{k>=1} x^(k^4))^n.

Extensions

More terms from Alois P. Heinz, Feb 04 2018

A307738 Number of partitions of n^3 into at most n cubes.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 4, 7, 18, 36, 66, 157, 329, 728, 1611, 3655, 8062, 18154, 40358, 89807, 199778, 444419, 984422, 2183461, 4827756, 10651083, 23465459, 51576034, 113092423, 247546849, 540538832, 1177836149, 2560897979, 5555722749, 12025952101, 25976048200
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 25 2019

Keywords

Comments

Does a(n+1) / a(n) ~ 2? - David A. Corneth, Sep 27 2019

Examples

			7^3 =
1^3 + 1^3 + 5^3 + 6^3 =
1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 =
1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3,
so a(7) = 4.

Links

Crossrefs

Cf. A000578, A025446-A025454, A030272, A105152, A259792, A298671, A298672, A307643, A307739.

Programs

  • PARI
    a(n) = {my(res = 0); res=aIterate(n^3, 1, n); res }
aIterate(s, m, q) = { if(s == 0, return(1)); if(q == 0, return(0)); sum(i = m, sqrtnint(s, 3), aIterate(s - i^3, i, q-1) ) } \\ David A. Corneth, Sep 23 2019

Extensions

a(21)-a(36) from David A. Corneth, Sep 23 2019

A298848 Number of partitions of n^3 into distinct cubes > 1.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 5, 4, 3, 4, 13, 11, 15, 20, 23, 34, 52, 49, 97, 118, 164, 192, 296, 330, 525, 745, 825, 1354, 1820, 1994, 3356, 4605, 5543, 8335, 12319, 13124, 21133, 28634, 33209, 51272, 71154, 85329, 126806, 174704, 210157, 310269, 433490, 511199
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 27 2018

Keywords

Examples

			a(6) = 2 because we have [216] and [125, 64, 27].

Links

Crossrefs

Cf. A000578, A003108, A030272, A078128, A259792, A279329, A280130, A298641, A298671, A298672.

Formula

a(n) = [x^(n^3)] Product_{k>=2} (1 + x^(k^3)).
a(n) = A280130(A000578(n)).

A298936 Number of ordered ways of writing n^2 as a sum of n nonnegative cubes.

Original entry on oeis.org

1, 1, 0, 6, 6, 20, 120, 7, 1689, 6636, 36540, 64020, 963996, 2894892, 19555965, 176079995, 955611188, 6684303780, 42462792168, 292378003753, 1886275214112, 13384059605364, 87399249887334, 624073002367892, 5080120229014734, 37587589611771480
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(3) = 6 because we have [8, 1, 0], [8, 0, 1], [1, 8, 0], [1, 0, 8], [0, 8, 1] and [0, 1, 8].

Links

Crossrefs

Cf. A000290, A000578, A023358, A030272, A218495, A259792, A291700, A298671, A298672, A298934, A298937.

Programs

  • Maple
    f:= n -> coeff(add(x^(k^3),k=0..floor(n^(2/3)))^n,x,n^2):
map(f, [$0..30]); # Robert Israel, Jan 29 2018
  • Mathematica
    Table[SeriesCoefficient[Sum[x^k^3, {k, 0, Floor[n^(2/3) + 1]}]^n, {x, 0, n^2}], {n, 0, 25}]

Formula

a(n) = [x^(n^2)] (Sum_{k>=0} x^(k^3))^n.

A298937 Number of ordered ways of writing n^2 as a sum of n positive cubes.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 7, 1, 0, 0, 9240, 34650, 1716, 48477, 551915, 6726720, 89973520, 102639744, 1824625081, 9915389400, 30143458884, 278196062760, 1995766236541, 6611689457736, 64547920386450, 236756174748626, 2315743488707806
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(7) = 7 because we have [8, 8, 8, 8, 8, 8, 1], [8, 8, 8, 8, 8, 1, 8], [8, 8, 8, 8, 1, 8, 8], [8, 8, 8, 1, 8, 8, 8], [8, 8, 1, 8, 8, 8, 8], [8, 1, 8, 8, 8, 8, 8] and [1, 8, 8, 8, 8, 8, 8].

Links

Crossrefs

Cf. A000290, A000578, A023358, A030272, A218495, A259792, A291700, A298671, A298672, A298934, A298936.

Programs

  • Mathematica
    Join[{1}, Table[SeriesCoefficient[Sum[x^k^3, {k, 1, Floor[n^(2/3) + 1]}]^n, {x, 0, n^2}], {n, 1, 27}]]

Formula

a(n) = [x^(n^2)] (Sum_{k>=1} x^(k^3))^n.

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)).
Showing 1-10 of 12 results. Next

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