A131689 -id:A131689 - cp's OEIS frontend

A133132 Number of surjections from an n-element set to a ten-element set.

3628800, 199584000, 6187104000, 142702560000, 2731586457600, 45950224320000, 703098107712000, 10009442963520000, 134672620008326400, 1732015476199008000, 21473732319740064000, 258323865658578720000

Offset: 10

Mohamed Bouhamida, Dec 16 2007

Vincenzo Librandi, Table of n, a(n) for n = 10..1000
Index entries for linear recurrences with constant coefficients, signature (55,-1320,18150,-157773,902055,-3416930,8409500,-12753576,10628640,-3628800).

Column k=10 of A131689.

Cf. A000918, A000919, A000920, A001117, A001118, A049435, A135456.

[10^n-10*9^n+45*8^n-120*7^n+210*6^n-252*5^n+210*4^n-120*3^n+45*2^n-10: n in [10..30]]; // Vincenzo Librandi, Apr 11 2012

With[{nn=30},Drop[CoefficientList[Series[(Exp[x]-1)^10,{x,0,nn}],x] Range[0,nn]!,10]] (* Harvey P. Dale, Sep 01 2016 *)

sum(k=1,10,(-1)^(10-k)*binomial(10,k)*k^n)

a(n) = 10^n-10*9^n+45*8^n-120*7^n+210*6^n-252*5^n+210*4^n-120*3^n+45*2^n-10.
a(n) = A049435(n) * 10!. - Max Alekseyev, Nov 13 2009
G.f.: 3628800*x^10/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)*(10*x-1)). - Colin Barker, Oct 25 2012
E.g.f.: (exp(x)-1)^10. - Alois P. Heinz, May 17 2016

More terms from Max Alekseyev, Nov 13 2009

Formula corrected by Charles R Greathouse IV, Mar 07 2010

A133360 Number of surjections from an n-element set to a nine-element set.

Original entry on oeis.org

362880, 16329600, 419126400, 8083152000, 130456085760, 1863435974400, 24359586451200, 297846188640000, 3457819037312640, 38528927611574400, 415357755774998400, 4358654246117808000, 44733116259693227520

Offset: 9

Mohamed Bouhamida, Dec 21 2007

Index entries for linear recurrences with constant coefficients, signature (45,-870,9450,-63273,269325,-723680,1172700,-1026576,362880).

Column k=9 of A131689.

Cf. A000918, A000919, A000920, A001117, A001118, A049447, A135456.

a(n) = Sum_{k=1..9} (-1)^(9-k)*binomial(9,k)*k^n.
a(n) = A049447(n) * 9!. - Max Alekseyev, Nov 12 2009
G.f.: -362880*x^9/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)). - Colin Barker, Oct 25 2012
E.g.f.: (exp(x) - 1)^9. - Ilya Gutkovskiy, Jun 19 2018

More terms from Max Alekseyev, Nov 12 2009

A194587 A triangle whose rows add up to the numerators of the Bernoulli numbers (with B(1) = 1/2). T(n, k) for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, -3, 4, 0, 1, -4, 3, 0, -15, 140, -270, 144, 0, 1, -20, 75, -96, 40, 0, -21, 868, -5670, 13104, -12600, 4320, 0, 1, -84, 903, -3360, 5600, -4320, 1260, 0, -15, 2540, -43470, 244944, -630000, 820800, -529200, 134400, 0, 1, -340, 9075, -74592, 278040, -544320, 582120, -322560, 72576

Offset: 0

Peter Luschny, Sep 17 2011

			[0] 1;
[1] 0,   1;
[2] 0,  -3,  4;
[3] 0,   1, -4,     3;
[4] 0, -15, 140, -270,    144;
[5] 0,   1, -20,   75,    -96,     40;
[6] 0, -21, 868, -5670, 13104, -12600,  4320;
[7] 0,   1, -84,   903, -3360,   5600, -4320, 1260;

Cf. A027641, A131689, A141056, A325871.

A194587 := proc(n, k) local i;
mul(i, i = select(isprime, map(i -> i + 1, numtheory[divisors](n)))):
(-1)^(n-k)*Stirling2(n, k) * k! / (k + 1): %%*% end:
seq(print(seq(A194587(n, k), k = 0..n)), n = 0..7);

T[n_, k_] := Times @@ Select[Divisors[n]+1, PrimeQ] (-1)^(n-k) StirlingS2[n, k]* k!/(k+1); Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 26 2019 *)

T(n, k) = (-1)^(n - k) * A131689(n, k) * A141056(n) / (k + 1).
Sum_{k=0..n} T(n, k) = A164555(n).
T(n, n) = A325871(n).

Edited by Peter Luschny, Jun 26 2019

Edited and flipped signs in odd indexed rows by Peter Luschny, Aug 20 2022

A285066 Triangle read by rows: T(n, m) = A285061(n, m)*m!, 0 <= m <= n.

Original entry on oeis.org

1, 1, 4, 1, 24, 32, 1, 124, 480, 384, 1, 624, 5312, 10752, 6144, 1, 3124, 52800, 203520, 276480, 122880, 1, 15624, 500192, 3279360, 7956480, 8110080, 2949120, 1, 78124, 4626720, 48633984, 187729920, 329441280, 268369920, 82575360, 1, 390624, 42265472, 687762432, 3969552384, 10672865280, 14615838720, 9909043200, 2642411520, 1, 1953124, 383514240, 9448097280, 78486589440, 303521218560, 621544734720, 696605736960, 404288962560, 95126814720

Offset: 0

Wolfdieter Lang, Apr 19 2017

This is the Sheffer triangle S2[4,1] = A285061 with column m scaled by m!. This is the fourth member of the triangle family A131689, A145901 and A284861.
This triangle appears in the o.g.f. G(n, x) = Sum_{m=0..n} T(n, m)*x^m/(1-x)^(m+1), n >= 0, of the power sequence {(1+4*m)^n}_{m >= 0}.
The diagonal sequence is A047053. The row sums give A285067. The alternating sum of row n is A141413(n+2), n >= 0.
The first column sequences are: A000012, 4*A003463, 2!*4^2*A016234.

			The triangle T(n, m) begins:
  n\m 0     1       2        3         4         5         6        7
  0:  1
  1:  1     4
  2:  1    24      32
  3:  1   124     480      384
  4:  1   624    5312    10752      6144
  5:  1  3124   52800   203520    276480    122880
  6:  1 15624  500192  3279360   7956480   8110080   2949120
  7:  1 78124 4626720 48633984 187729920 329441280 268369920 82575360
  ...
row 8: 1 390624 42265472 687762432 3969552384 10672865280 14615838720 9909043200 2642411520
row 9: 1 1953124 383514240 9448097280 78486589440 303521218560 621544734720 696605736960 404288962560 95126814720
...

Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.

Cf. A000012, 4*A003463, 32*A016234, A111578 (T(n, m)/4^m), A131689, A141413, A145901, A225118, A225472, A225473, A284861, A285061.

T[n_, m_]:=Sum[Binomial[m, k]*(-1)^(k - m)*(1 + 4k)^n, {k, 0, n}]; Table[T[n, m], {n, 0, 10},{m, 0, n}] // Flatten (* Indranil Ghosh, May 02 2017 *)

from sympy import binomial
def T(n, m):
    return sum([binomial(m, k)*(-1)**(k - m)*(1 + 4*k)**n for k in range(n + 1)])
for n in range(21):
    print([T(n, m) for m in range(n + 1)])
# Indranil Ghosh, May 02 2017

T(n, m) = A285061(n, m)*m! = A111578(n, m)*(4^m*m!), 0 <= m <= n.
T(n, m) = Sum_{k=0..n} binomial(m,k)*(-1)^(k-m)*(1+4*k)^n.
T(n, m) = Sum_{j=0..n} binomial(n-j,m-j)*A225118(n,n-j).
Recurrence: T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m and T(n, m) =
4*m*T(n-1, m-1) + (1+4*m)*T(n-1, m) for n >= 1, m=0..n.
E.g.f. row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)/(1 - x*(exp(4*z) - 1)).
E.g.f. column m: exp(x)*(exp(4*x) - 1)^m, m >= 0.
O.g.f. column m: m!*(4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x), m >= 0.

A300729 Number of arrangements on a line of n finite closed intervals (possibly of zero length) with k distinct endpoints (n >= 1, 1 <= k <= 2*n).

Original entry on oeis.org

1, 1, 1, 7, 12, 6, 1, 25, 138, 294, 270, 90, 1, 79, 1056, 5298, 12780, 16020, 10080, 2520, 1, 241, 7050, 70350, 334710, 875970, 1335600, 1184400, 567000, 113400, 1, 727, 44472, 817746, 6849900, 31500180, 87348240, 152643960, 169533000, 116235000, 44906400, 7484400

Offset: 1

Peter Bala, Mar 12 2018

A122193(n,k) equals the number of arrangements on a line of n (nondegenerate) finite closed intervals having k distinct endpoints. The entries T(n,k) of the present table satisfy T(n,k) = A122193(n,k) + A122193(n,k+1). Proof. In an arrangement contributing to T(n,k) either the intervals are all nondegenerate, and there are A122193(n,k) arrangements of this type, or at least one of the intervals in the arrangement is degenerate. The following argument to show there are A122193(n,k+1) arrangements of the latter type is taken from the solution to the problem posed in the 'IBM Ponder This' link.
In an arrangement of n nondegenerate finite closed intervals having k+1 distinct endpoints, the rightmost point is the right endpoint of one or more intervals. If we move each of these right endpoints to coincide with their corresponding left endpoint then we obtain an arrangement of n finite closed intervals with k distinct endpoints, where at least one of the intervals has zero length. The reverse mapping is clear: given an arrangement of n finite closed intervals with k distinct endpoints, where at least one of the intervals has zero length, take each interval of zero length and move all the right endpoints of these degenerate intervals to a single new rightmost point. This produces an arrangement of n nondegenerate finite closed intervals having k+1 distinct endpoints. (End proof)
Most of the properties of the present table now follow from the properties of A122193.
Reading the table by antidiagonals produces A059515.

			Table begins
      |k=0   1   2     3     4      5      6      7     8
---------------------------------------------------------
  n=0 |  1
    1 |  0   1   1
    2 |  0   1   7    12     6
    3 |  0   1  25   138   294    270     90
    4 |  0   1  79  1056  5298  12780  16020  10080  2520
   ...
T(2,3) = 12: The 12 arrangements with 3 endpoints of two (possibly degenerate) intervals [a, A] and [b, B] are
     aA-b-B, b-aA-B, b-B-aA, bB-a-A, a-bB-A, a-A-bB,
     ab-A-B, ab-B-A, a-b-AB, b-a-AB, a-bA-B, b-a-AB.
Here, for example, the notation aA-b-B indicates a = A < b < B, so the interval [a, A] is degenerate and lies to the left of the nondegenerate interval [b, B].
Row 2: (1, 7, 12, 6)
(x*(x + 1)/2)^2 = C(x,1) + 7*C(x,2) + 12*C(x,3) + 6*C(x,4).
Row 3: (1, 25, 138, 294, 270, 90)
(x*(x + 1)/2)^3 = C(x,1) + 25*C(x,2) + 138*C(x,3) + 294*C(x,4) + 270*C(x,5) + 90*C(x,6).
Sums of powers of triangular numbers:
Sum_{i = 1..n-1} (i*(i+1)/2)^2 = C(n,2) + 7*C(n,3) + 12*C(n,4) + 6*C(n,5);
Sum_{i = 1..n-1} (i*(i+1)/2)^3 = C(n,2) + 25*C(n,3) + 138*C(n,4) + 294*C(n,5) + 270*C(n,6) + 90*C(n,7).

Peter Bala, Deformations of the Hadamard product of power series
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
IBM Ponder This, Jan 01 2001
Helmut Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Cf. A059516 (row sums), A059515, A087127, A122193, A131689.

A300729 := proc (n, k)
add((-1)^(k-i)*binomial(k, i)*((1/2)*i*(i+1))^n, i = 0..k);
end proc:
for n from 0 to 8 do
seq(A300729(n, k), k = 0..2*n)
end do;

T[0, 0] = 1; T[n_, k_] := Sum[(-1)^(k-i)*Binomial[k, i]*((1/2)*i*(i+1))^n, {i, 0, k}]; Table[T[n, k], {n, 1, 6}, {k, 1, 2 n}] // Flatten (* Jean-François Alcover, Mar 16 2018 *)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * (i*(i+1)/2)^n for 0 <= k <= 2*n.
T(n,k) = A122193(n,k) + A122193(n,k+1).
T(n,k) = k*(k+1)/2*T(n-1,k) + k^2*T(n-1,k-1) + k*(k-1)/2*T(n-1,k-2) for 1 < k <= 2*n with boundary conditions T(0,0) = 1, T(0,n) = 0 for n >= 1; T(n,1) = 1 for n >= 1 and T(n,k) = 0 if k > 2*n.
Double e.g.f.: exp(-x)*Sum_{n>=0} exp( binomial(n+1,2)*y )* x^n/n! = 1 + (x + x^2/2!)*y + (x + 7*x^2/2! + 12*x^3/3! + 6*x^4/4!)*y^2/2! + ....
The n-th row of the table is given by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318 and v_n is the sequence (0, 1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Cf. A087127 and A122193.
n-th row polynomial R(n,x) = (x + x^2) o ... o (x + x^2) (n factors) where o denotes the black diamond product of power series as defined by Dukes and White.
R(n,x) = Sum_{i >= 1} (i*(i+1)/2)^n*x^i/(1 + x)^(i+1) for n >= 1.
x*R(n,x) = (1 + x)* the n-th row polynomial of A122193 for n >= 1.
(1 + x)*R(n,x) = x * the n-th row polynomial of A087127 for n >= 1.
Sum_{k = 1..2*n} T(n,k)*binomial(x,k) = (binomial(x+1,2))^n for n >= 1.
Sum_{i = 1..n-1} (i*(i+1)/2)^m = Sum_{k = 1..2*m} T(m,k)*binomial(n,k+1) for m >= 1. See Example section below.
R(n,x) = 1/2^n*Sum_{k = 0..n} binomial(n,k)*F(n+k,x), where F(n,x) = Sum_{k = 0..n} k!*Stirling2(n,k)*x^k is the n-th Fubini polynomial, the n-th row polynomial of A131689.
R(n+1,x) = 1/2*(x + x^2) * (d/dx)^2 ( (x + x^2)*R(n,x) ).
R(n,x) = R(n,-1 - x).
The zeros of R(n,x) belong to the interval [-1, 0].
For n >= 1, the alternating sum of the n-th row equals 0.
E.g.f. as a continued fraction: 1/(1 + (x + x^2)*(1 - exp(t))/(1 + (x + x^2)*(1 -exp(2*t))/(1 + (x + x^2)*(1 - exp(3*t))/(1 + ...)))) = 1 + (x + x^2)*t + (x + 7*x^2 + 12*x^3 + 6*x^4)*t^2/2! + ... (use Prodinger, equation 1.1). - Peter Bala, Jun 13 2019

A323099 Number T(n,k) of colored set partitions of [n] where exactly k colors are used for the elements; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 2, 4, 0, 5, 30, 30, 0, 15, 210, 540, 360, 0, 52, 1560, 7800, 12480, 6240, 0, 203, 12586, 109620, 316680, 365400, 146160, 0, 877, 110502, 1583862, 7366800, 14733600, 13260240, 4420080, 0, 4140, 1051560, 23995440, 169011360, 521640000, 792892800, 584236800, 166924800

Offset: 0

Alois P. Heinz, Aug 30 2019

			Triangle T(n,k) begins:
  1;
  0,   1;
  0,   2,      4;
  0,   5,     30,      30;
  0,  15,    210,     540,     360;
  0,  52,   1560,    7800,   12480,     6240;
  0, 203,  12586,  109620,  316680,   365400,   146160;
  0, 877, 110502, 1583862, 7366800, 14733600, 13260240, 4420080;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Partition of a set

Columns k=0-1 give: A000007, A000110 (for n>0).
Row sums give A121017.
Main diagonal gives A137341.
Cf. A000142, A008277, A048993, A019538, A131689.

A:= proc(n, k) option remember; `if`(n=0, 1, add(
      A(n-j, k)*binomial(n-1, j-1)*k^j, j=1..n))
    end:
T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);
# second Maple program:
T:= (n, k)-> combinat[bell](n)*Stirling2(n,k)*k!:
seq(seq(T(n, k), k=0..n), n=0..10);

A[n_, k_] := A[n, k] = If[n==0, 1, Sum[A[n-j, k] Binomial[n-1, j-1] k^j, {j, 1, n}]];
T[n_, k_] := Sum[A[n, k - i] (-1)^i Binomial[k, i], {i, 0, k}];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten
(* second program: *)
T[n_, k_] := BellB[n] StirlingS2[n, k] k!;
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Dec 08 2020, after Alois P. Heinz *)

T(n,k) = Bell(n) * Sum_{i=0..k} (k-i)^n * (-1)^i * C(k,i).
T(n,k) = Bell(n) * A131689(n,k).
T(n,k) = Bell(n) * Stirling2(n,k) * k!.

A326477 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 2 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 4, 3, 0, 46, 60, 15, 0, 1114, 1848, 840, 105, 0, 46246, 88770, 54180, 12600, 945, 0, 2933074, 6235548, 4574130, 1469160, 207900, 10395, 0, 263817646, 605964450, 505915410, 199849650, 39729690, 3783780, 135135

Offset: 0

Peter Luschny, Jul 08 2019

			Triangle starts:
[0] [1]
[1] [0, 1]
[2] [0, 4, 3]
[3] [0, 46, 60, 15]
[4] [0, 1114, 1848, 840, 105]
[5] [0, 46246, 88770, 54180, 12600, 945]
[6] [0, 2933074, 6235548, 4574130, 1469160, 207900, 10395]

Row sums A094088. Alternating row sums A153881 starting at 0.
Main diagonal A001147. Associated set partitions A241171.
A129062 (m=1, associated with A131689), this sequence (m=2), A326587 (m=3, associated with A278073), A326585 (m=4, associated with A278074).

CL := f -> PolynomialTools:-CoefficientList(f, x):
FL := s -> ListTools:-Flatten(s, 1):
StirPochConv := proc(m, n) local P, L; P := proc(m, n) option remember;
`if`(n = 0, 1, add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n)) end:
L := CL(P(m, n)); CL(expand(add(L[k+1]*pochhammer(x,k)/k!, k=0..n))) end:
FL([seq(StirPochConv(2,n), n = 0..7)]);

P[, 0] = 1; P[m, n_] := P[m, n] = Sum[Binomial[m*n, m*k]*P[m, n-k]*x, {k, 1, n}] // Expand;
T[m_][n_] := CoefficientList[P[m, n], x].Table[Pochhammer[x, k]/k!, {k, 0, n}] // CoefficientList[#, x]&;
Table[T[2][n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jul 21 2019 *)

def StirPochConv(m, n):
    z = var('z'); R = ZZ[x]
    F = [i/m for i in (1..m-1)]
    H = hypergeometric([], F, (z/m)^m)
    P = R(factorial(m*n)*taylor(exp(x*(H-1)), z, 0, m*n + 1).coefficient(z, m*n))
    L = P.list()
    S = sum(L[k]*rising_factorial(x,k) for k in (0..n))
    return expand(S).list()
for n in (0..6): print(StirPochConv(2, n))

For m >= 1 let P(m,0) = 1 and P(m, n) = Sum_{k=1..n} binomial(m*n, m*k)*P(m, n-k)*x for n > 0. Then T_{m}(n, k) = Sum_{k=0..n} ([x^k]P(m, n))*rf(x,k)/k! where rf(x,k) are the rising factorial powers. T(n, k) = T_{2}(n, k).

A326585 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 4 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 36, 35, 0, 12046, 17820, 5775, 0, 16674906, 30263480, 16216200, 2627625, 0, 65544211366, 135417565890, 93516348900, 26189163000, 2546168625, 0, 588586227465426, 1334168329550300, 1083314031995250, 402794176785000, 69571511509500, 4509264634875

Offset: 0

Peter Luschny, Jul 21 2019

			Triangle starts:
[0] [1]
[1] [0, 1]
[2] [0, 36, 35]
[3] [0, 12046, 17820, 5775]
[4] [0, 16674906, 30263480, 16216200, 2627625]
[5] [0, 65544211366, 135417565890, 93516348900, 26189163000, 2546168625]
[6] [0, 588586227465426, 1334168329550300, 1083314031995250, 402794176785000, 69571511509500, 4509264634875]

Row sums A243665. Main diagonal A025036.

A129062 (m=1, associated with A131689), A326477 (m=2, associated with A241171), A326587 (m=3, associated with A278073), this sequence (m=4, associated with A278074).

Maple
```
# See A326477.
```

T(n, k) = T_{4}(n, k) where T_{m}(n, k) is defined in A326477.

A326587 Coefficients of polynomials related to ordered set partitions. Triangle read by rows, T_{m}(n, k) for m = 3 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 11, 10, 0, 645, 924, 280, 0, 111563, 197802, 101640, 15400, 0, 42567981, 86271640, 57717660, 15415400, 1401400, 0, 30342678923, 67630651098, 53492240256, 19158419280, 3144741600, 190590400

Offset: 0

Peter Luschny, Jul 20 2019

			Triangle starts:
0 [1]
1 [0, 1]
2 [0, 11, 10]
3 [0, 645, 924, 280]
4 [0, 111563, 197802, 101640, 15400]
5 [0, 42567981, 86271640, 57717660, 15415400, 1401400]
6 [0, 30342678923, 67630651098, 53492240256, 19158419280, 3144741600, 190590400]

Row sums A243664. Main diagonal A025035.

A129062 (m=1, associated with A131689), A326477 (m=2, associated with A241171), this sequence (m=3, associated with A278073), A326585 (m=4, associated with A278074).

Maple
```
# See A326477.
```

T(n, k) = T_{3}(n, k) where T_{m}(n, k) is defined in A326477.

A371898 Triangle read by rows: T(n, k) = n * k * (T(n-1, k-1) + T(n-1, k)) for k > 0 with initial values T(n, 0) = 1 and T(i, j) = 0 for j > i.

Original entry on oeis.org

1, 1, 1, 1, 4, 4, 1, 15, 48, 36, 1, 64, 504, 1008, 576, 1, 325, 5680, 22680, 31680, 14400, 1, 1956, 72060, 510480, 1304640, 1382400, 518400, 1, 13699, 1036224, 12233340, 50823360, 94046400, 79833600, 25401600, 1, 109600, 16798768, 318469536, 2017814400, 5794790400, 8346240000, 5893171200, 1625702400

Offset: 0

Werner Schulte, Apr 11 2024

			Lower triangular array starts:
n\k :  0      1        2         3         4         5         6         7
==========================================================================
  0 :  1
  1 :  1      1
  2 :  1      4        4
  3 :  1     15       48        36
  4 :  1     64      504      1008       576
  5 :  1    325     5680     22680     31680     14400
  6 :  1   1956    72060    510480   1304640   1382400    518400
  7 :  1  13699  1036224  12233340  50823360  94046400  79833600  25401600
  etc.

Cf. A000012 (column 0), A007526 (column 1), A001044 (main diagonal).

Cf. A000166, A048993, A131689, A320031, A371766, A371767.

T[n_, k_] := Sum[(-1)^(k - j)*Binomial[k, j]*HypergeometricPFQ[{1, -n}, {}, -j], {j, 0, k}];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten  (* Peter Luschny, Apr 12 2024 *)

T(n, k) = if(k==0, 1, if(k > n, 0, n*k*(T(n-1, k-1) + T(n-1, k))))

T(n, k) = Sum_{i=k..n} A131689(i, k) * n! / (n-i)!.
T(n, k) = n! * k! * (Sum_{i=0..n-k} A048993(n-i, k) / i!).
T(n, k) = Sum_{i=0..k} (-1)^(k-i) * binomial(k, i) * A320031(n, i).
Conjecture: E.g.f. of column k is exp(t) * t^k * k! / (Prod_{i=0..k} (1 - i*t)).
Conjecture: Sum_{k=0..n} (-1)^(n-k) * T(n, k) = A000166(n).
T(n, k) = A371766(n, k) * A371767(n, k). - Peter Luschny, Apr 14 2024

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