A092271 Triangle read by rows. First in a series of triangular arrays counting permutations of partitions.

1, 1, 1, 2, 3, 1, 6, 8, 6, 1, 24, 30, 20, 10, 1, 120, 144, 90, 40, 15, 1, 720, 840, 504, 210, 70, 21, 1, 5040, 5760, 3360, 1344, 420, 112, 28, 1, 40320, 45360, 25920, 10080, 3024, 756, 168, 36, 1, 362880, 403200, 226800, 86400, 25200, 6048, 1260, 240, 45, 1, 3628800, 3991680, 2217600, 831600, 237600, 55440, 11088, 1980, 330, 55, 1

Offset

1,4

Comments

Generate signatures in accordance with A086141. Map to partitions in accordance with A025487. Calculate the number of permutations in accordance with Abramowitz and Stegun, p. 831 (reference M2). Display the results as illustrated by A090774. The second array is:

3

20 15

90 120 45

504 630 420 105

...

Apart from the main diagonal, appears to be the same as A211603 (see also A238363) and is related to the infinitesimal generator of A008290; if so, the off-diagonal triangle entries are given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2. - Peter Bala, Feb 13 2017

Let aut(p) denote the size of the centralizer of the partition p (see A339016 for the definition). Then row(n) = [n!/aut(p) for p in P], where P are the partitions of n with largest part k and length n + 1 - k. Row sums are A121726. - Peter Luschny, Nov 19 2020

References

Abramowitz and Stegun, p. 831.

Links

Table of n, a(n) for n=1..66.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Example

The triangle begins:
1:    1
2:    1   1
3:    2   3  1
4:    6   8  6  1
5:   24  30 20 10  1
6:  120 144 90 40 15 1
  ...
From Peter Luschny, Nov 19 2020: (Start):
The combinatorial interpretation is illustrated by this computation of row 6:
6! / aut([6])                = 720 / A339033(6, 1) = 720/6   = 120 = T(6, 1)
6! / aut([5, 1])             = 720 / A339033(6, 2) = 720/5   = 144 = T(6, 2)
6! / aut([4, 1, 1])          = 720 / A339033(6, 3) = 720/8   =  90 = T(6, 3)
6! / aut([3, 1, 1, 1])       = 720 / A339033(6, 4) = 720/18  =  40 = T(6, 4)
6! / aut([2, 1, 1, 1, 1])    = 720 / A339033(6, 5) = 720/48  =  15 = T(6, 5)
6! / aut([1, 1, 1, 1, 1, 1]) = 720 / A339033(6, 6) = 720/720 =   1 = T(6, 6)
-------------------------------------------------------------------------------
                                                         Sum:  410 = A121726(6)
(End)

Mathematica

f[list_] :=Total[list]!/Apply[Times, list]/Apply[Times, Map[Length, Split[list]]!]; Table[Append[Map[f, Select[Partitions[n], Count[#, Except[1]] == 1 &]], 1], {n, 1, 10}] // Grid (* Geoffrey Critzer, Nov 07 2015 *)

Prog

(SageMath)
def A092271(n, k):
    if n == k: return 1
    return factorial(n) // ((n + 1 - k)*factorial(k - 1))
for n in (1..9): print(n, [A092271(n, k) for k in (1..n)])
def A092271Row(n):
    if n == 0: return [1]
    f = factorial(n); S = []
    for k in range(n, 0, -1):
        for p in Partitions(n, max_part=k, inner=[k], length=n+1-k):
            S.append(f // p.aut())
    return S
for n in (1..9): print(A092271Row(n)) # Peter Luschny, Nov 20 2020

Crossrefs

Cf. A007290, A025487, A086141, A090774, A008290, A111492, A211603, A238363, A121726, A339016, A339033.

Sequence in context: A036039 A324254 A279038 * A054115 A100822 A347766

Adjacent sequences: A092268 A092269 A092270 * A092272 A092273 A092274

Keyword

nonn,tabl

Author

Alford Arnold, Feb 14 2004

Extensions

More terms from Geoffrey Critzer, Nov 10 2015

Status

approved