A339033 -id:A339033 - cp's OEIS frontend

A339034 Row sums of A339033.

1, 1, 4, 11, 35, 147, 805, 5399, 42273, 375787, 3728261, 40788255, 487539769, 6319430483, 88272658797, 1321745733511, 21117967813025, 358591883025339, 6448525343069653, 122424951294889967, 2446864618294774281, 51354975368171586595, 1129258990476358286909

Offset: 0

Peter Luschny, Nov 20 2020

nonn

Paolo Xausa, Table of n, a(n) for n = 0..440

Cf. A339033.

A339034[n_] := If[n == 0, 1, n! + Sum[(n+1-k)*(k-1)!, {k, n-1}]];
Array[A339034, 25, 0] (* Paolo Xausa, Jan 31 2024 *)

a(n) = if (n==0, 1, n! - (n-1)! + sum(k=1, n, (n+1-k)*(k-1)!)); \\ Michel Marcus, Dec 02 2020

def A339034(n):
    if n == 0: return 1
    d = factorial(n) - factorial(n - 1)
    return add((n + 1 - k)*factorial(k - 1) for k in (1..n)) + d
print([A339034(n) for n in (0..22)])

a(n) = n! - (n - 1)! + Sum_{k=1..n} (n + 1 - k)*(k - 1)! for n > 0.

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

Original entry on oeis.org

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

Alford Arnold, Feb 14 2004

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

			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)

Abramowitz and Stegun, p. 831.

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].

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

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 *)

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

More terms from Geoffrey Critzer, Nov 10 2015

A121726 Sum sequence A000522 then subtract 0,1,2,3,4,5,...

Original entry on oeis.org

1, 2, 6, 21, 85, 410, 2366, 16065, 125665, 1112074, 10976174, 119481285, 1421542629, 18348340114, 255323504918, 3809950976993, 60683990530209, 1027542662934898, 18430998766219318, 349096664728623317, 6962409983976703317, 145841989688186383338, 3201192743180799343822

Offset: 1

Alford Arnold, Aug 17 2006

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

			A000522 begins     1 2 5 16 65 326 ...
with sums          1 3 8 24 89 415 ...
so sequence begins 1 2 6 21 85 410 ...
.
From _Peter Luschny_, Nov 19 2020: (Start):
The combinatorial interpretation is illustrated by this computation of a(5):
5! / aut([5])             = 120 / A339033(5, 1) = 120/5   = 24
5! / aut([4, 1])          = 120 / A339033(5, 2) = 120/4   = 30
5! / aut([3, 1, 1])       = 120 / A339033(5, 3) = 120/6   = 20
5! / aut([2, 1, 1, 1])    = 120 / A339033(5, 4) = 120/12  = 10
5! / aut([1, 1, 1, 1, 1]) = 120 / A339033(5, 5) = 120/120 =  1
--------------------------------------------------------------
                                                Sum: a(5) = 85
(End)

Also the row sums of A092271.

Cf. A000522, A006231, A002104, A339016, A339033.

f[list_] :=Total[list]!/Apply[Times, list]/Apply[Times, Map[Length, Split[list]]!]; Table[Total[Map[f, Select[Partitions[n], Count[#, Except[1]] == 1 &]]] + 1, {n, 1, 20}] (* Geoffrey Critzer, Nov 07 2015 *)

A000522(n)={ return( sum(k=0,n,n!/k!)) ; } A121726(n)={ return(sum(k=0,n-1,A000522(k))-n+1) ; } { for(n=1,25, print1(A121726(n),",") ; ) ; } \\ R. J. Mathar, Sep 02 2006

def A121726(n):
    def h(n, k):
        if n == k: return 1
        return factorial(n)//((n + 1 - k)*factorial(k - 1))
    return sum(h(n, k) for k in (1..n))
print([A121726(n) for n in (1..23)])
# Demonstrates the combinatorial view:
def A121726(n):
    if n == 0: return 1
    f = factorial(n); S = 0
    for k in (0..n):
        for p in Partitions(n, max_part=k, inner=[k], length=n+1-k):
            S += (f // p.aut())
    return S
print([A121726(n) for n in (1..23)]) # Peter Luschny, Nov 20 2020

a(n) = A006231(n) + 1 = A002104(n) - (n-1). - Franklin T. Adams-Watters, Aug 29 2006

E.g.f.: exp(x)*(log(1/(1-x)) - x + 1). - Geoffrey Critzer, Nov 07 2015

More terms from Franklin T. Adams-Watters, Aug 29 2006

More terms from R. J. Mathar, Sep 02 2006

A339016 A classification of permutations based on their cycle length and the size of the centralizer of their cycle type. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 0, 6, 0, 0, 0, 3, 21, 0, 0, 0, 0, 35, 85, 0, 0, 0, 0, 55, 255, 410, 0, 0, 0, 0, 0, 1015, 1659, 2366, 0, 0, 0, 0, 0, 2485, 10528, 11242, 16065, 0, 0, 0, 0, 0, 2240, 58149, 92064, 84762, 125665, 0, 0, 0, 0, 0, 0, 228221, 760725, 805530, 722250, 1112074

Offset: 0

Peter Luschny, Nov 19 2020

The size of the centralizer of a partition p is aut(p) = Product_{j = 1..k} m(j)!*j^m(j), where m(j) is the multiplicity of j as a part of p. (For instance p = [2, 2, 2] -> aut(p) = 3!*2^3.)
Let M be the matrix with M(k, r) = Sum_{p in P(n, k)} n! / aut(p) where P(n, k) are the partitions of n with largest part k and length(p) = r. Then T(n, k) = Sum_{j=0..k} M(j, k-j+1), which are the antidiagonal sums of the upper triangular part of the matrix M.
In the example section below it is explained how the matrix M leads to a two-dimensional classification of the permutations of [n] which project to the unsigned Stirling cycle numbers and the number of permutations with longest cycle length.

			Triangle starts:
0:  [1]
1:  [0, 1]
2:  [0, 0, 2]
3:  [0, 0, 0, 6]
4:  [0, 0, 0, 3,  21]
5:  [0, 0, 0, 0,  35,   85]
6:  [0, 0, 0, 0,  55,  255,     410]
7:  [0, 0, 0, 0,   0, 1015,    1659,  2366]
8:  [0, 0, 0, 0,   0, 2485,   10528, 11242, 16065]
9:  [0, 0, 0, 0,   0, 2240,   58149, 92064, 84762, 125665]
----------------------------------------------------------
Sum  1, 1, 2, 9, 111, 6080, 2331767, ...
.
Examples for the basic two-dimensional classification of permutations (dots indicate zeros):
.
* Case n = 6:
   |   1     2     3    4    5    6  | Sum
-------------------------------------|----
1  |   .     .     .    .    .   [1] |   1
2  |   .     .   [ 15] [45] [15]     |  75
3  |   .   [ 40] [120] [40]          | 200
4  |   .   [ 90] [ 90]               | 180
5  |   .   [144]                     | 144
6  | [120]                           | 120
-------------------------------------|----
Sum| 120,  274,   225,  85,  15,  1  | 720
.
Antidiagonals: [40 + 15, 90 + 120 + 45, 120 + 144 + 90 + 40 + 15 + 1]
Leads to row 6 (disregarding leading zeros): 55 + 255 + 410 = 720.
.
* Case n = 7:
   |  1      2     3     4     5    6    7  | Sum
--------------------------------------------|-----
1  |  .      .     .     .     .    .   [1] |    1
2  |  .      .     .   [105] [105] [21]     |  231
3  |  .      .   [490] [420] [ 70]          |  980
4  |  .    [420] [630] [210]                | 1260
5  |  .    [504] [504]                      | 1008
6  |  .    [840]                            |  840
7  | [720]                                  |  720
--------------------------------------------|-----
Sum| 720,  1764,  1624, 735,  175,  21,  1  | 5040
.
Antidiagonals: [420+490+105, 504+630+420+105, 720+840+504+210+70+21+1]
Leads to row 7 (disregarding leading zeros): 1015 + 1659 + 2366 = 5040
.
* Column sums of the matrix give the unsigned Stirling cycle numbers, A132393.
* Row sums of the matrix give the number of permutations of n elements whose longest cycle have length k, A126074.
* The main antidiagonal of the matrix gives the number of n-permutations that are pure cycles of length n - k, A092271.
* The entries of the matrix sum to n!. In particular the sum over all row sums, the sum over all column sums, and the sum over all antidiagonal sums is n!.
* The columns of the triangle are finite in the sense that their entries become ultimately zero. Column sums of the triangle are A339015.

Cf. A000142 (row sums), A339015 (column sums), A132393, A126074, A092271, A121726, A339033, A006231, A002104.

# For illustration computes also A132393 and A126074 (remove the #).
def A339016Row(n):
    f = factorial(n); M = matrix(n + 2)
    for k in (0..n):
        for p in Partitions(n, max_part=k, inner=[k]):
            M[k, len(p)] += (f // p.aut())
    # print("max cyc len", [sum(M[k, j] for j in (0..n+1)) for k in (0..n)])
    # print("Stirling 1 ", [sum(M[j, k] for j in (0..n+1)) for k in (0..n)])
    if n == 0: return [1]
    return [sum(M[j, k-j+1] for j in srange(k, 0, -1)) for k in (0..n)]
for n in (0..9): print(A339016Row(n))

T(n, n) = A006231(n) + 1 = A002104(n) - (n-1) (after Franklin T. Adams-Watters in A121726).

cp's OEIS Frontend

A339034 Row sums of A339033.

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

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A121726 Sum sequence A000522 then subtract 0,1,2,3,4,5,...

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A339016 A classification of permutations based on their cycle length and the size of the centralizer of their cycle type. Triangle read by rows, T(n, k) for 0 <= k <= n.

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