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 10 results.

A020919 Partition numbers mod 11.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 0, 4, 0, 8, 9, 1, 0, 2, 3, 0, 0, 0, 0, 6, 0, 0, 1, 1, 2, 0, 5, 7, 0, 0, 5, 0, 0, 1, 1, 0, 3, 0, 0, 0, 4, 0, 0, 0, 1, 1, 2, 3, 5, 0, 0, 0, 0, 8, 0, 1, 0, 2, 0, 0, 7, 0, 0, 6, 0, 9, 1, 1, 2, 3, 5, 7, 0, 4, 0, 0, 9, 1, 0, 2, 3, 0, 0, 0, 0, 0, 8, 0, 1, 1, 2, 3, 5, 7, 0, 4, 0, 8, 0, 1, 1, 2, 3, 5, 0
Offset: 0

Views

Author

Keywords

Comments

Zeros are uncommonly dense at beginning of sequence, no other modulus exhibits this behavior.

Crossrefs

Programs

  • Mathematica
    Mod[PartitionsP[Range[0,120]],11] (* Harvey P. Dale, May 14 2023 *)
  • PARI
    a(n) = numbpart(n) % 11; \\ Michel Marcus, Jul 14 2022

Formula

a(n) = A000041(n) mod 11. - Sean A. Irvine, May 04 2019

A087183 Partition numbers of the form 3*k.

Original entry on oeis.org

3, 15, 30, 42, 135, 231, 297, 627, 792, 1002, 1575, 2436, 5604, 8349, 10143, 14883, 31185, 37338, 44583, 63261, 105558, 147273, 239943, 281589, 329931, 614154, 1121505, 1505499, 3087735, 4087968, 4697205, 8118264, 15796476, 44108109
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Comments

The numbers m such that 3 divides A000041(m) are given in A083214. Klarreich writes: no one has proved whether there are infinitely many partition numbers divisible by 3, although it is known that there are infinitely many partition numbers divisible by 2. - Jonathan Vos Post, Jul 31 2008
Intersection of A008585 and A000041. - Reinhard Zumkeller, Nov 03 2009

References

  • Erica Klarreich, Pieces of numbers: a proof brings closure to a dramatic tale of partitions and primes, Science News, Jun 18 2005.

Crossrefs

Programs

Formula

A079978(a(n))*A167392(a(n)) = 1. - Reinhard Zumkeller, Nov 03 2009
a(n) = 3*A213365(n). - Omar E. Pol, May 08 2013
a(n) = A000041(A083214(n)). - Amiram Eldar, May 22 2025

A068908 Number of partitions of n modulo 5.

Original entry on oeis.org

1, 1, 2, 3, 0, 2, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 1, 2, 0, 0, 2, 2, 2, 0, 0, 3, 1, 0, 3, 0, 4, 2, 4, 3, 0, 3, 2, 2, 0, 0, 3, 3, 4, 1, 0, 4, 3, 4, 3, 0, 1, 3, 4, 1, 0, 1, 3, 4, 0, 0, 2, 0, 1, 4, 0, 3, 0, 4, 0, 0, 3, 0, 3, 4, 0, 4, 1, 3, 4, 0, 1, 2, 0, 4, 0, 2, 2, 3, 4, 0, 3, 4, 2, 2, 0, 4, 4, 0, 1, 0, 2, 1, 4, 0, 0
Offset: 0

Views

Author

Henry Bottomley, Mar 05 2002

Keywords

Comments

Of the partitions of numbers from 1 to 100000: 36256 are 0, 15758 are 1, 16133 are 2, 16028 are 3 and 15825 are 4 modulo 5, largely because the number of partitions of 5m+4 is always a multiple of 5.

Crossrefs

Programs

  • Mathematica
    Mod[PartitionsP[Range[0,110]],5] (* Harvey P. Dale, Dec 20 2023 *)
  • PARI
    a(n) = numbpart(n) % 5; \\ Michel Marcus, Jul 14 2022

Formula

a(n) = A010874(A000041(n)) = A068906(5, n).
a(n) = Pm(n,1) with Pm(n,k) = if kReinhard Zumkeller, Jun 09 2009]

A087180 Number partition numbers <= P(n) of the form 3*k (P = A000041).

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 4, 4, 4, 4, 5, 5, 6, 7, 7, 7, 8, 9, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 14, 15, 15, 16, 16, 16, 16, 17, 18, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 23, 24, 25, 25, 25, 25, 26, 26, 26, 26, 27, 27, 28, 28, 28, 28, 28, 29, 29, 30, 31, 31, 31, 31, 32
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Boole[Mod[PartitionsP[n], 3] == 0], {n, 0, 100}]] (* Amiram Eldar, May 22 2025 *)

Formula

a(n) + A087181(n) + A087182(n) = n + 1.

A087181 Number partition numbers <= P(n) of the form 3*k+1 (P = A000041).

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 9, 9, 9, 9, 9, 9, 10, 10, 11, 12, 12, 12, 12, 12, 12, 12, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 17, 17, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 22, 22, 22, 22, 23, 23, 23
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Boole[Mod[PartitionsP[n], 3] == 1], {n, 0, 100}]] (* Amiram Eldar, May 22 2025 *)

Formula

A087180(n) + a(n) + A087182(n) = n + 1.

A087182 Number partition numbers <= P(n) of the form 3*k+2 (P = A000041).

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 17, 17, 17, 18, 19, 20, 20, 21, 21, 21, 21, 21, 22, 22, 23, 23, 24
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Boole[Mod[PartitionsP[n], 3] == 2], {n, 0, 100}]] (* Amiram Eldar, May 22 2025 *)

Formula

A087180(n) + A087181(n) + a(n) = n + 1.

A087184 Partition numbers of the form 3*k+1.

Original entry on oeis.org

1, 1, 7, 22, 385, 490, 1255, 3010, 3718, 12310, 17977, 21637, 75175, 89134, 204226, 386155, 451276, 831820, 1300156, 1741630, 5392783, 6185689, 10619863, 18004327, 20506255, 34262962, 49995925, 64112359, 104651419, 150198136
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Select[PartitionsP[Range[0, 100]], Mod[#, 3] == 1 &] (* Amiram Eldar, May 22 2025 *)

Formula

a(n) = A000041(A237276(n)). - Amiram Eldar, May 22 2025

A087185 Partition numbers of the form 3*k+2.

Original entry on oeis.org

2, 5, 11, 56, 77, 101, 176, 1958, 4565, 6842, 26015, 53174, 124754, 173525, 526823, 715220, 966467, 2012558, 2323520, 2679689, 3554345, 7089500, 9289091, 12132164, 13848650, 23338469, 26543660, 30167357, 38887673, 56634173
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 23 2003

Keywords

Crossrefs

Programs

  • Mathematica
    Select[PartitionsP[Range[200]],Divisible[#-2,3]&] (* Harvey P. Dale, Apr 22 2016 *)

Formula

a(n) = A000041(A237277(n)). - Amiram Eldar, May 22 2025

A068906 Square array read by ascending antidiagonals of partitions of k modulo n.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 1, 1, 3, 2, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 0, 0, 3, 2, 5, 3, 2, 1, 0, 0, 1, 3, 1, 1, 5, 3, 2, 1, 0, 0, 0, 2, 0, 5, 0, 5, 3, 2, 1, 0, 0, 0, 2, 2, 3, 4, 7, 5, 3, 2, 1, 0, 1, 2, 2, 0, 4, 1, 3, 7, 5, 3, 2, 1, 0, 1, 2, 0, 2, 0, 1, 7, 2, 7, 5, 3, 2, 1
Offset: 1

Views

Author

Henry Bottomley, Mar 05 2002

Keywords

Comments

0 is disproportionately common modulo 5, 7 and 11, largely because T(5,5m+4)=T(7,7m+5)=T(11,11m+6)=0.

Examples

			Rows start 0,0,0,0,0,...; 1,0,1,1,1,...; 1,2,0,2,1,...; 1,2,3,1,3,...; 1,2,3,0,2,1,...; 1,2,3,5,1,5,...; 1,2,3,5,0,...; 1,2,3,5,7,...; etc.
		

Crossrefs

Rows 2, 3, 5, 7 and 11 give A040051, A068907, A068908, A068909, A020919.

Formula

T(n, k) =A051127(n, A000041(k))

A068909 Number of partitions of n modulo 7.

Original entry on oeis.org

1, 1, 2, 3, 5, 0, 4, 1, 1, 2, 0, 0, 0, 3, 2, 1, 0, 3, 0, 0, 4, 1, 1, 2, 0, 5, 0, 0, 1, 1, 4, 3, 5, 0, 4, 1, 1, 0, 3, 0, 0, 0, 2, 2, 2, 3, 5, 0, 0, 2, 1, 4, 0, 0, 0, 0, 3, 2, 2, 3, 5, 0, 4, 2, 2, 2, 3, 5, 0, 4, 3, 2, 4, 6, 5, 0, 0, 2, 2, 4, 3, 5, 0, 0, 3, 3, 6, 6, 3, 0, 1, 3, 3, 4, 3, 5, 0, 0, 4, 3, 4, 6, 5, 0, 1
Offset: 0

Views

Author

Henry Bottomley, Mar 05 2002

Keywords

Comments

Of the partitions of numbers from 1 to 100000: 27193 are 0, 12078 are 1, 12203 are 2, 12260 are 3, 12231 are 4, 12003 are 5 and 12032 are 6 modulo 7, largely because the number of partitions of 7m+5 is always a multiple of 7.

Crossrefs

Programs

  • Mathematica
    Table[Mod[PartitionsP[n],7],{n,0,110}] (* Harvey P. Dale, Feb 17 2018 *)
  • PARI
    a(n) = numbpart(n) % 7; \\ Michel Marcus, Jul 14 2022

Formula

a(n) = A010876(A000041(n)) = A068906(7, n).
a(n) = Pm(n,1) with Pm(n,k) = if kReinhard Zumkeller, Jun 09 2009]
Showing 1-10 of 10 results.