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 1-3 of 3 results.

A124774 Multinomial coefficients for compositions in standard order.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 3, 6, 1, 4, 6, 12, 4, 12, 12, 24, 1, 5, 10, 20, 10, 30, 30, 60, 5, 20, 30, 60, 20, 60, 60, 120, 1, 6, 15, 30, 20, 60, 60, 120, 15, 60, 90, 180, 60, 180, 180, 360, 6, 30, 60, 120, 60, 180, 180, 360, 30, 120, 180, 360, 120, 360, 360, 720, 1, 7, 21, 42, 35, 105
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
Number of ways to distribute labeled objects into boxes, with the number of objects in each box being specified by the composition.

Examples

			Composition number 11 is 2,1,1; there are 6 choices for the pair of objects in the first box, then 2 choices for the object in the next box, so a(11) = 6*2 = 12.
The table starts:
1
1
1 2
1 3 3 6
		

Crossrefs

Cf. A066099, A124773, A011782 (row lengths), A000670 (row sums), A036039.

Formula

For composition b(1),...,b(k), a(n) = (Sum_{i=1}^k b(i))! / (Product_{i=1}^k b(i)!).

A124772 Number of set partitions associated with compositions in standard order.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 1, 2, 1, 1, 1, 4, 6, 6, 4, 8, 4, 4, 1, 3, 3, 3, 1, 2, 1, 1, 1, 5, 10, 10, 10, 20, 10, 10, 5, 15, 15, 15, 5, 10, 5, 5, 1, 4, 6, 6, 4, 8, 4, 4, 1, 3, 3, 3, 1, 2, 1, 1, 1, 6, 15, 15, 20, 40, 20, 20, 15, 45, 45, 45, 15, 30, 15, 15, 6, 24, 36, 36, 24, 48, 24, 24, 6, 18
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
Arrange the parts of the set partition by the smallest member of each part and read off the part sizes. E.g., for 1|24|3, the associated composition is 1,2,1. When the set partition is presented as the sequence of parts that each member is in, simply count the times each part number occurs. This representation for 1|24|3 is {1,2,3,2}.

Examples

			Composition number 11 is 2,1,1; the associated set partitions are 12|3|4, 13|2|4 and 14|2|3, so a(11) = 3.
The table starts:
1
1
1 1
1 2 1 1
		

Crossrefs

Cf. A066099, A124773, A011782 (row lengths), A000110 (row sums), A036040, A080575.

Formula

For composition b(1),...,b(k), a(n) = Product_{i=1}^k C((Sum_{j=i}^k b(j))-1, b(i)-1).

A306549 a(n) is the product of the positions of the zeros in the binary expansion of n (the most significant bit having position 1).

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 3, 1, 24, 6, 8, 2, 12, 3, 4, 1, 120, 24, 30, 6, 40, 8, 10, 2, 60, 12, 15, 3, 20, 4, 5, 1, 720, 120, 144, 24, 180, 30, 36, 6, 240, 40, 48, 8, 60, 10, 12, 2, 360, 60, 72, 12, 90, 15, 18, 3, 120, 20, 24, 4, 30, 5, 6, 1, 5040, 720, 840, 120, 1008
Offset: 0

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Author

Rémy Sigrist, May 04 2019

Keywords

Comments

Apparently, the variant where the least significant bit has position 1 corresponds to A124773.

Examples

			The first terms, alongside the positions of zeros and the binary representation of n, are:
  n   a(n)  Pos.zeros  bin(n)
  --  ----  ---------  ------
   0     1  {1}             0
   1     1  {}              1
   2     2  {2}            10
   3     1  {}             11
   4     6  {2,3}         100
   5     2  {2}           101
   6     3  {3}           110
   7     1  {}            111
   8    24  {2,3,4}      1000
   9     6  {2,3}        1001
  10     8  {2,4}        1010
  11     2  {2}          1011
  12    12  {3,4}        1100
  13     3  {3}          1101
  14     4  {4}          1110
  15     1  {}           1111
		

Crossrefs

Programs

  • Mathematica
    A306549[n_] := Times @@ Flatten[Position[IntegerDigits[n, 2], 0]];
    Array[A306549, 100, 0] (* Paolo Xausa, Jun 01 2024 *)
  • PARI
    a(n) = my (b=binary(n)); prod(k=1, #b, if (b[k]==0, k, 1))
    
  • PARI
    a(n) = vecprod(Vec(select(x->(x==0), binary(n), 1)));
    
  • Python
    from math import prod
    def a(n): return prod(i for i, bi in enumerate(bin(n)[2:], 1) if bi == "0")
    print([a(n) for n in range(70)]) # Michael S. Branicky, Jun 01 2024

Formula

a(n) = A070939(n)! / A306286(n).
a(2*n) = a(n) * (1+A070939(n)).
a(2*n+1) = a(n).
a(2^k) = (k+1)! for any k >= 0.
a(2^k-1) = 1 for any k >= 0.
a(2^k-2) = k for any k >= 1.
Showing 1-3 of 3 results.