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.

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A295341 The number of partitions of n in which at least one part is a multiple of 3.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 4, 6, 9, 14, 20, 29, 41, 57, 78, 106, 142, 189, 250, 327, 425, 549, 705, 900, 1144, 1445, 1819, 2279, 2844, 3534, 4379, 5403, 6648, 8152, 9969, 12152, 14780, 17920, 21682, 26163, 31504, 37842, 45371, 54270, 64800, 77211, 91842, 109031, 129235, 152897
Offset: 0

Views

Author

R. J. Mathar, Nov 20 2017

Keywords

Comments

From Gus Wiseman, May 23 2022: (Start)
Also the number of integer partitions of n with at least one part appearing more than twice. The Heinz numbers of these partitions are given by A046099. For example, the a(0) = 0 though a(8) = 9 partitions are:
. . . (111) (1111) (2111) (222) (2221) (2222)
(11111) (3111) (4111) (5111)
(21111) (22111) (22211)
(111111) (31111) (32111)
(211111) (41111)
(1111111) (221111)
(311111)
(2111111)
(11111111)
(End)

Examples

			From _Gus Wiseman_, May 23 2022: (Start)
The a(0) = 0 through a(8) = 9 partitions with a part that is a multiple of 3:
  .  .  .  (3)  (31)  (32)   (6)     (43)     (53)
                      (311)  (33)    (61)     (62)
                             (321)   (322)    (332)
                             (3111)  (331)    (431)
                                     (3211)   (611)
                                     (31111)  (3221)
                                              (3311)
                                              (32111)
                                              (311111)
(End)
		

Crossrefs

The complement is counted by A000726, ranked by A004709.
These partitions are ranked by A354235.
This is column k = 3 of A354234.
For 2 instead of 3 we have A047967, ranked by A013929 and A324929.
For 4 instead of 3 we have A295342, ranked by A046101.
A000041 counts integer partitions, strict A000009.
A046099 lists non-cubefree numbers.

Programs

Formula

a(n) = A000041(n)-A000726(n).

A354234 Triangle read by rows where T(n,k) is the number of integer partitions of n with at least one part divisible by k.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 5, 3, 1, 1, 7, 4, 2, 1, 1, 11, 7, 4, 2, 1, 1, 15, 10, 6, 3, 2, 1, 1, 22, 16, 9, 6, 3, 2, 1, 1, 30, 22, 14, 8, 5, 3, 2, 1, 1, 42, 32, 20, 13, 8, 5, 3, 2, 1, 1, 56, 44, 29, 18, 12, 7, 5, 3, 2, 1, 1, 77, 62, 41, 27, 17, 12, 7, 5, 3, 2, 1, 1
Offset: 1

Views

Author

Gus Wiseman, May 22 2022

Keywords

Comments

Also partitions of n with at least one part appearing k or more times. It would be interesting to have a bijective proof of this.

Examples

			Triangle begins:
   1
   2  1
   3  1  1
   5  3  1  1
   7  4  2  1  1
  11  7  4  2  1  1
  15 10  6  3  2  1  1
  22 16  9  6  3  2  1  1
  30 22 14  8  5  3  2  1  1
  42 32 20 13  8  5  3  2  1  1
  56 44 29 18 12  7  5  3  2  1  1
  77 62 41 27 17 12  7  5  3  2  1  1
For example, row n = 5 counts the following partitions:
  (5)      (32)    (32)   (41)  (5)
  (32)     (41)    (311)
  (41)     (221)
  (221)    (2111)
  (311)
  (2111)
  (11111)
At least one part appearing k or more times:
  (5)      (221)    (2111)   (11111)  (11111)
  (32)     (311)    (11111)
  (41)     (2111)
  (221)    (11111)
  (311)
  (2111)
  (11111)
		

Crossrefs

The complement is counted by A061199.
Differences of consecutive terms are A091602.
Column k = 1 is A000041.
Column k = 2 is A047967, ranked by A013929 and A324929.
Column k = 3 is A295341, ranked by A046099 and A354235.
Column k = 4 is A295342.
A000041 counts integer partitions, strict A000009.
A047966 counts uniform partitions.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],MemberQ[#/k,_?IntegerQ]&]],{n,1,15},{k,1,n}]
    - or -
    Table[Length[Select[IntegerPartitions[n],Max@@Length/@Split[#]>=k&]],{n,1,15},{k,1,n}]
  • PARI
    \\ here P(k,n) is partitions with no part divisible by k as g.f.
    P(k,n)={1/prod(i=1, n, 1 - if(i%k, x^i) + O(x*x^n))}
    M(n,m=n)={my(p=P(n+1,n)); Mat(vector(m, k, Col(p-P(k,n), -n) ))}
    { my(A=M(12)); for(n=1, #A, print(A[n,1..n])) } \\ Andrew Howroyd, Jan 19 2023
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