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 1-6 of 6 results.

A068413 a(n) = number of partitions of 2^n.

Original entry on oeis.org

1, 2, 5, 22, 231, 8349, 1741630, 4351078600, 365749566870782, 4453575699570940947378, 61847822068260244309086870983975, 18116048323611252751541173214616030020513022685, 6927233917602120527467409170319882882996950147283323368445315320451
Offset: 0

Views

Author

Henry Bottomley, Mar 03 2002

Keywords

Examples

			a(2)=5 since there are 5 partitions of 2^2=4: 4, 3+1, 2+2, 2+1+1, 1+1+1+1+1.

Links

Crossrefs

Cf. A000041, A000079, A018819, A067735.

Programs

  • Mathematica
    Table[ PartitionsP[2^n], {n, 0, 12}]

Formula

a(n) = A000041(A000079(n)).
a(n) ~ exp(Pi*sqrt(2^(n+1)/3))/(sqrt(3)*2^(n+2)). - Ilya Gutkovskiy, Jan 13 2017

A237999 Number of partitions of 2^n into parts that are at most n with at least one part of each size.

Original entry on oeis.org

0, 1, 1, 2, 9, 119, 4935, 596763, 211517867, 224663223092, 734961197081208, 7614278809664610952, 256261752606028225485183, 28642174350851846128820426827, 10830277060032417592098008847162727, 14068379226083299071248895931891435683229
Offset: 0

Views

Author

Alois P. Heinz, Feb 16 2014

Keywords

Comments

From Gus Wiseman, May 31 2019: (Start)
Also the number of strict integer partitions of 2^n with n parts. For example, the a(1) = 1 through a(4) = 9 partitions are (A = 10):
(2) (31) (431) (6532)
(521) (6541)
(7432)
(7531)
(7621)
(8431)
(8521)
(9421)
(A321)
(End)

Examples

			a(1) = 1: 11.
a(2) = 1: 211.
a(3) = 2: 3221, 32111.
a(4) = 9: 433321, 443221, 4322221, 4332211, 4432111, 43222111, 43321111, 432211111, 4321111111.

Links

Crossrefs

Column k=2 of A238012.
Cf. A236810, A237512, A237998, A238000, A238001.
Cf. A000009, A002033, A067735, A126796, A283111.

Programs

  • Mathematica
    a[n_] := SeriesCoefficient[Product[1/(1 - x^j), {j, 1, n}], {x, 0, 2^n - n*(n + 1)/2}];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 15}] (* Jean-François Alcover, Aug 19 2018 *)

Formula

a(n) = [x^(2^n-n*(n+1)/2)] Product_{j=1..n} 1/(1-x^j).
a(n) ~ 2^(n*(n-1)) / (n!*(n-1)!). - Vaclav Kotesovec, Jun 05 2015

A069878 Number of partitions of 10^n into distinct parts.

Original entry on oeis.org

1, 10, 444793, 8635565795744155161506, 1122606574548038398976040173670530159089991444775125551802871247408332723840
Offset: 0

Views

Author

Robert G. Wilson v, May 03 2002

Keywords

Links

Crossrefs

Row 10 of A347621.
Cf. A000009, A011557, A064682, A067735 & A070177.

Programs

  • Mathematica
    Table[ PartitionsQ[10^n], {n, 0, 4}]
  • PARI
    a(n) = polcoef(prod(k=1, 10^n, 1+x^k+x*O(x^(10^n))), 10^n); \\ Seiichi Manyama, Sep 10 2021

Formula

a(n) = A000009(A011557(n)). - Michel Marcus, Sep 10 2021

A347621 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) is the number of partitions of n^k into distinct parts.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 6, 8, 2, 1, 1, 1, 32, 192, 32, 3, 1, 1, 1, 390, 84756, 16444, 142, 4, 1, 1, 1, 16444, 5807301632, 11784471548, 3207086, 668, 5, 1, 1, 1, 4013544, 2496696209705056142, 16816734263788624008200, 74443865946867656, 1258238720, 3264, 6, 1
Offset: 0

Views

Author

Seiichi Manyama, Sep 09 2021

Keywords

Examples

			Square array begins:
  1, 1,  1,     1,           1, ...
  1, 1,  1,     1,           1, ...
  1, 1,  2,     6,          32, ...
  1, 2,  8,   192,       84756, ...
  1, 2, 32, 16444, 11784471548, ...

Crossrefs

Columns k=0..3 give A000012, A000009, A072243, A281501.
Rows n=0+1, 2-3 give A000012, A067735, A070235.
Main diagonal gives A064682.
Cf. A347615, A347630.

Programs

  • Mathematica
    Table[If[n == k == 0, 1, PartitionsQ[#^k] &[n - k]], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Sep 09 2021 *)
  • PARI
    T(n, k) = polcoef(prod(j=1, n^k, 1+x^j+x*O(x^(n^k))), n^k);

Formula

T(n,k) = A000009(n^k).

A070235 Number of partitions of 3^n into distinct terms.

Original entry on oeis.org

1, 2, 8, 192, 84756, 5807301632, 2496696209705056142, 4077067982967062771640042697374910, 1661834856564765736512658856496529945178404778001420955780288
Offset: 0

Views

Author

Robert G. Wilson v, May 08 2002

Keywords

Crossrefs

Cf. A000009, A067735, A069878, A248728.

Programs

  • Mathematica
    Table[ PartitionsQ[3^n], {n, 0, 10}]

Formula

a(n) ~ exp(Pi*sqrt(3^(n-1)))/(4*3^(3*n/4+1/4)). - Ilya Gutkovskiy, Jan 13 2017

A079281 Number of compositions of 2^n into distinct parts.

Original entry on oeis.org

1, 1, 3, 19, 435, 74875, 348317763, 294729601581739, 682404222981720262704195, 298417646219775679438413815505895285915, 13661663328896434876017827688479176004409461863714010289523203
Offset: 0

Views

Author

Henry Bottomley, Feb 08 2003

Keywords

Examples

			a(2) = 3 since the compositions of 2^2=4 into distinct parts are 4, 3+1 and 1+3.

Links

Crossrefs

Cf. A058891 (offset for compositions of 2^n), A067735, A068413.

Programs

  • Maple
    b:= proc(n, i) option remember; local m; m:= i*(i+1)/2;
      `if`(n=m, x^i, `if`(n>m, 0,
       expand(b(n, i-1)+`if`(i>n, 0, x*b(n-i, i-1)))))
    end:
a:= n->(p->add(coeff(p, x, i)*i!, i=0..degree(p)))(b(2^n$2)):
seq(a(n), n=0..9); # Alois P. Heinz, Apr 27 2014
  • Mathematica
    b[n_, i_] := b[n, i] = With[{ m = i*(i+1)/2}, If[n==m, x^i, If[n>m, 0, Expand[b[n, i-1] + If[i>n, 0, x*b[n-i, i-1]]]]]]; a[n_] := Function[{p}, Sum[Coefficient[p, x, i]*i!, {i, 0, Exponent[p, x]}]][b[2^n, 2^n]]; Table[a[n], {n, 0, 9}] (* Jean-François Alcover, Oct 05 2015, after Alois P. Heinz *)

Formula

a(n) = A032020(A000079(n)).
Showing 1-6 of 6 results.

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