A130534 -id:A130534 - cp's OEIS frontend

A260196 1, -3, followed by -1's.

1, -3, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1

Offset: 0

Paul Curtz, Jul 19 2015

1/(n+1) is the inverse Akiyama-Tanigawa transform of A164555(n)/A027642(n).
For more on the Akiyama-Tanigawa transform, see Links (correction: page 7 read here A164555 instead of A027641) and A177427.
Here:
   1,  -3, -1, -1, -1, -1, ...
   4,  -4,  0,  0,  0,  0, ...
   8,  -8,  0,  0,  0,  0, ...
  16, -16,  0,  0,  0,  0, ...
  etc.
Other process, using signed A130534(n), different of A008275(n):
1,                            1/1,                                      1,
1, 4,                      (  1,     -1)/1,                            -3,
1, 4, 8,                   (  2,     -3,   1)/2,                       -1,
1, 4, 8, 16,          *    (  6,    -11,   6,   -1)/6,             =   -1,
1, 4, 8, 16, 32,           ( 24,    -50,  35,  -10,  1)/24,            -1,
1, 4, 8, 16, 32, 64,       (120,   -274, 225,  -85, 15, -1)/120,       -1,
etc.                       etc.                                        etc.
Via the modified Stirling numbers of the first kind, the second triangle, Iw(n), is the inverse of Worpitzky transform A163626(n).
a(n) is the third sequence of a family beginning with
1,  1,  1,  1,  1,  1,  1,  1, ...              = A000012(n)
1,  0,  0,  0,  0,  0,  0,  0,  0, ...          = A000007(n)
1, -3, -1, -1, -1, -1, -1, -1, -1, -1, ... .
A000012(n) is the inverse Akiyama-Tanigawa transform of A000007(n), with or without its second term.
A000007(n) is the inverse Akiyama-Tanigawa transform of A000012(n), with or without its second term.
a(n) is the inverse Akiyama-Tanigawa transform of 2^n omitting the second term i.e. 2.

Colin Barker, Table of n, a(n) for n = 0..1000
Masanobu Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, Journal of Integer Sequences, 3(2000), article 00.2.9
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A000007, A000012, A008275, A027642, A130534, A151821, A163626, A164555, A177427.

first(m)=vector(m,i,i--;if(i>1,-1,if(i==0,1,if(i==1,-3)))) \\ Anders Hellström, Aug 28 2015

Vec(-(2*x^2-4*x+1)/(x-1) + O(x^100)) \\ Colin Barker, Sep 11 2015

Inverse Akiyama-Tanigawa transform of A151821(n).
From Colin Barker, Sep 11 2015: (Start)
a(n) = -1 for n>1.
a(n) = a(n-1) for n>2.
G.f.: -(2*x^2-4*x+1) / (x-1).
(End)

A358622 Regular triangle read by rows. T(n, k) = [[n, k]], where [[n, k]] are the second order Stirling cycle numbers (or second order reciprocal Stirling numbers). T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 6, 3, 0, 0, 0, 24, 20, 0, 0, 0, 0, 120, 130, 15, 0, 0, 0, 0, 720, 924, 210, 0, 0, 0, 0, 0, 5040, 7308, 2380, 105, 0, 0, 0, 0, 0, 40320, 64224, 26432, 2520, 0, 0, 0, 0, 0, 0, 362880, 623376, 303660, 44100, 945, 0, 0, 0, 0, 0

Offset: 0

Peter Luschny, Nov 23 2022

[[n, k]] are the number of permutations of an n-set having at least two elements in each orbit. These permutations have no fixed points and therefore [[n, k]] is the number of k-orbit derangements of an n-set. This is the definition and notation (doubling the stacked delimiters of the Stirling cycle numbers) as given by Fekete (see link).
The formal definition expresses the second order Stirling cycle numbers as a binomial sum over second order Eulerian numbers (see the first formula below). The terminology 'associated Stirling numbers of first kind' used elsewhere should be dropped in favor of the more systematic one used here.
Also the Bell transform of the factorial numbers with 0! = 0. For the definition of the Bell transform see A264428.

			Triangle T(n, k) starts:
[0] 1;
[1] 0,     0;
[2] 0,     1,     0;
[3] 0,     2,     0,     0;
[4] 0,     6,     3,     0,    0;
[5] 0,    24,    20,     0,    0,  0;
[6] 0,   120,   130,    15,    0,  0,  0;
[7] 0,   720,   924,   210,    0,  0,  0,  0;
[8] 0,  5040,  7308,  2380,  105,  0,  0,  0,  0;
[9] 0, 40320, 64224, 26432, 2520,  0,  0,  0,  0,  0;

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994, thirty-fourth printing 2022.

Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 8.
Antal E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.

A008306 is an irregular subtriangle with more information.
Cf. A000166 (row sums), A024000 (alternating row sums).
Cf. A130534, A201637, A341101, A264428, A269940.

P := (n, x) -> (-x)^n*hypergeom([-n, x], [], 1/x):
row := n -> seq(coeff(simplify(P(n, x)), x, k), k = 0..n):
for n from 0 to 9 do row(n) od;
# Alternative:
T := (n, k) -> add(binomial(n, k - j)*abs(Stirling1(n - k + j, j))*(-1)^(k - j), j =  0..k): for n from 0 to 9 do seq(T(n, k), k = 0..n) od;
# Using the e.g.f.:
egf := (exp(t)*(1 - t))^(-z): ser := series(egf, t, 12):
seq(print(seq(n!*coeff(coeff(ser, t, n), z, k), k=0..n)), n = 0..9);
# Using second order Eulerian numbers:
A358622 := proc(n, k) local j;
add(binomial(j, n - 2*k)*combinat:-eulerian2(n - k, j), j = 0..n-k) end:
seq(seq(A358622(n, k), k = 0..n), n = 0..12);
# Using generalized Laguerre polynomials:
P := (n, x) -> (-1)^n*n!*LaguerreL(n, -n - x, -x):
row := n -> seq(coeff(simplify(P(n, x)), x, k), k = 0..n):
seq(print(row(n)), n = 0..9);

# recursion over rows
from functools import cache
@cache
def StirlingCycleOrd2(n: int) -> list[int]:
    if n == 0: return [1]
    if n == 1: return [0, 0]
    rov: list[int] = StirlingCycleOrd2(n - 2)
    row: list[int] = StirlingCycleOrd2(n - 1) + [0]
    for k in range(1, n // 2 + 1):
        row[k] = (n - 1) * (rov[k - 1] + row[k])
    return row
for n in range(9): print(StirlingCycleOrd2(n))
# Alternative, using function BellMatrix from A264428.
from math import factorial
def f(k: int) -> int:
    return factorial(k) if k > 0 else 0
print(BellMatrix(f, 9))

T(n, k) = Sum_{j=0..n-k} binomial(j, n - 2*k)*<>, where <> denote the second order Eulerian numbers (extending Knuth's notation).
T(n, k) = [x^n] (-x)^n * hypergeom([-n, x], [], -1/x).
T(n, k) = n!*[z^k][t^n] (exp(t)*(1 - t))^(-z). (Compare with (exp(t)/(1 - t))^z, which is the e.g.f. of the Sylvester polynomials A341101.)
T(n, k) = [x^k] (-1)^n * n! * L(n, -x - n, -x), where L(n, a, x) is the n-th generalized Laguerre polynomial.
T(n, k) = Sum_{j=0..k} binomial(n, k - j)*[n - k + j, j]*(-1)^(k - j), where [n, k] denotes the (signless) Stirling cycle numbers.
T(n, k) = (n - 1) * (T(n-2, k-1) + T(n-1, k)) with suitable boundary conditions.
T(n + k, k) = A269940(n, k), which might be called the Ward cycle numbers.

A370518 Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n} binomial(n,i)(n-i)!Stirling1(i,k)TC(n,i) where TC(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)Stirling2(i+3,i+1)*(-1)^(i) for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, -5, 1, 14, -9, 1, -18, 29, -12, 1, 0, -22, 35, -14, 1, 0, -26, 15, 25, -15, 1, 0, -60, 4, 75, -5, -15, 1, 0, -204, -56, 259, 70, -56, -14, 1, 0, -912, -484, 1092, 609, -168, -126, -12, 1, 0, -5040, -3708, 5480, 4599, -231, -882, -210, -9, 1, 0, -33120, -30024, 31820, 36350, 3675, -6027, -2370, -300, -5, 1

Offset: 0

Igor Victorovich Statsenko, Feb 21 2024

Generalized Stirling numbers of the first kind of the second order.

			n\k     0     1     2     3     4     5     6
0:      1
1:     -5     1
2:     14    -9     1
3:    -18    29   -12     1
4:      0   -22    35   -14     1
5:      0   -26    15    25   -15     1
6:      0   -60     4    75    -5   -15     1

Igor Victorovich Statsenko, On the ordinal numbers of triangles of generalized special numbers, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.

For m=0 the formula gives the sequence A130534; for m=1 the formula gives the sequence A094645. In this case, we assume that A130534 consists of generalized Stirling numbers of the first kind of zero order, and A094645 consists of generalized Stirling numbers of the first kind of the first order.

C:=(n,k)->n!/(k!*(n-k)!) : T0:=(m,n,k)->sum(C(n+1,n-k-p)*Stirling2(p+m+1,p+1)*((-1)^p), p=0..n-k) : T:=(m,n,k)->sum(C(n,r)*(n-r)!*Stirling1(r,k)*T0(m,n,r), r=0..n)  m:=2 : seq(seq T(m,n,k), k=0..n), n=0..10);

T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(m,n,i) where TC(m,n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^(i),m = 2 for n >= 0.

A387152 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} binomial(k, j)*|Stirling1(n, j)|.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 2, 3, 3, 1, 0, 6, 7, 6, 4, 1, 0, 24, 23, 16, 10, 5, 1, 0, 120, 98, 57, 30, 15, 6, 1, 0, 720, 514, 257, 115, 50, 21, 7, 1, 0, 5040, 3204, 1407, 546, 205, 77, 28, 8, 1, 0, 40320, 23148, 9076, 3109, 1021, 336, 112, 36, 9, 1

Offset: 0

Peter Luschny, Aug 27 2025

			Array begins:
  [0]  1,     1,      1,      1,       1,       1,       1, ...
  [1]  0,     1,      2,      3,       4,       5,       6, ...
  [2]  0,     1,      3,      6,      10,      15,      21, ...
  [3]  0,     2,      7,     16,      30,      50,      77, ...
  [4]  0,     6,     23,     57,     115,     205,     336, ...
  [5]  0,    24,     98,    257,     546,    1021,    1750, ...
  [6]  0,   120,    514,   1407,    3109,    6030,   10696, ...
  [7]  0,   720,   3204,   9076,   20695,   41330,   75356, ...
  [8]  0,  5040,  23148,  67456,  157865,  323005,  602517, ...
  [9]  0, 40320, 190224, 567836, 1358564, 2837549, 5396650, ...

Rows: A000012 [0], A001477 [1], A000217 [2], A005581 [3], A387204 [4].
Columns: A000007 [0], A000142 [shifted, 1], A387205 [2].
Contains A271700 in transpose.
Cf. A211210 (main diagonal), A130534.

A := (n, k) -> add(binomial(k, j)*abs(Stirling1(n, j)), j = 0..n):
seq(seq(A(n-k, k), k = 0..n), n = 0..10);
# Expanding rows or columns:
RowSer := n -> series((1+x)^k*GAMMA(x + n)/GAMMA(x), x, 12):
Trow := n -> k -> coeff(RowSer(n), x, k):
ColSer := n -> series(orthopoly:-L(n, log(1 - x)), x, 12):
Tcol := k -> n -> n! * coeff(ColSer(k), x, n):
seq(lprint(seq(Trow(n)(k), k = 0..7)), n = 0..9);
seq(lprint(seq(Tcol(k)(n), n = 0..7)), k = 0..9);

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if n == 0: return 1
    if k == 0: return 0
    return (n - 1) * T(n - 1, k) + T(n, k - 1) - (n - 2) * T(n - 1, k - 1)
for n in range(7): print([T(n, k) for k in range(7)])

T(n, k) = n! * [x^n] Laguerre(k, log(1 - x)).
From Natalia L. Skirrow, Aug 27 2025: (Start)
D-finite with T(n,k) = (n-1)*T(n-1,k)+T(n,k-1)-(n-2)*T(n-1,k-1).
O.g.f.: hypergeom([1,y/(1-y)],[],x)/(1-y).
Row o.g.f.: (y/(1-y))_n/(1-y), where (x)_n is the Pochhammer symbol/rising factorial.
Row o.g.f. is also 0^n + y/(1-y)^(n+1)*Prod_{j=1..n-2}(j+1-j*y).
E.g.f.: 1/((1-y)*(1-x)^(y/(1-y))).
Column e.g.f.: hypergeom([-k],[1],log(1-y)).
T(n,k) = [x^k] (1+x)^k*(x)_n.
(End)

A387205 a(n) = (n - 1)!*(2 + Harmonic(n - 1)) if n >= 1, and a(0) = 1.

Original entry on oeis.org

1, 2, 3, 7, 23, 98, 514, 3204, 23148, 190224, 1752336, 17886240, 200377440, 2444446080, 32256800640, 457822229760, 6954511737600, 112579862169600, 1934780446771200, 35181735469977600, 674855347635302400, 13618752053114880000, 288426695123589120000, 6396478234890670080000

Offset: 0

Peter Luschny, Aug 27 2025

Paolo Xausa, Table of n, a(n) for n = 0..400

Cf. A387152 (column 2), A001008, A130534.

a := n -> if n = 0 then 1 else (n-1)!*(2 + harmonic(n-1)) fi:
ser := series(LaguerreL(2, log(1 - x)), x, 24): a := n -> n! * coeff(ser, x, n):
seq(a(n), n = 0..23);

A387205[n_] := If[n == 0, 1, (n - 1)!*(2 + HarmonicNumber[n - 1])];
Array[A387205, 25, 0] (* Paolo Xausa, Aug 29 2025 *)

a(n) = if (n>0, (n-1)!*(2 + sum(i=1, n-1, 1/i)), 1); \\ Michel Marcus, Aug 27 2025

a(n) = 2*|Stirling1(n, 1)| + |Stirling1(n, 2)| for n >= 1.
a(n) = n! * [x^n] Laguerre(2, log(1 - x)).
a(n) = Gamma(n)*(PolyGamma(n) + EulerGamma + 2) for n >= 1.
Conjecture: Maple returns the exponential series expansion at x = 0:
a(n) = n! * [x^n] (1 + tau + (log(x - 1)^2 - (tau + 4)*log(x - 1) - Pi^2)/2) where tau = 2*Pi*I.

A059418 Triangle T(n,k) arising from enumeration of permutations with ordered orbits, read by rows (1<=k<=n).

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 12, 7, 4, 1, 60, 33, 19, 7, 1, 360, 192, 109, 47, 11, 1, 2520, 1320, 737, 344, 102, 16, 1, 20160, 10440, 5742, 2801, 956, 198, 22, 1, 181440, 93240, 50634, 25349, 9493, 2342, 352, 29, 1, 1814400, 927360, 498312, 253426, 101293, 28229

Offset: 1

N. J. A. Sloane, Jan 30 2001

From Peter Bala, Jul 08 2012: (Start)
A non-plane recursive tree is a rooted labeled plane tree (the children of a node are not ordered) with the property that the labels increase along any path from the root to a leaf. T(n,k) counts the total number of vertices of outdegree k among the set of n! non-plane recursive trees on n+1 vertices. An example is given below.
(End)

			1; 1,1; 3,2,1; 12,7,4,1; 60,33,19,7,1; ...
Row 3: [12,7,4,1]. There are 6 non-plane recursive trees on 4 nodes.
The total number of nodes of outdegree 0 = 1+2+2+2+2+3 = 12;
The total number of nodes of outdegree 1 = 3+1+1+1+1 = 7;
The total number of nodes of outdegree 2 = 1+1+1+1 = 4;
The total number of nodes of outdegree 3 = 1;
...................................................................
.0o......0o..........0o..........0o.........0o...........0o........
..|.......|........../.\........./.\......../.\........../|\.......
..|.......|........./...\......./...\....../...\......../.|.\......
.1o......1o.......1o.....o3...1o....o2...2o.....o1...../..|..\.....
..|....../.\.......|...........|..........|..........1o..2o...o3...
..|...../...\......|...........|..........|........................
.2o...2o.....o3...2o..........3o.........3o........................
..|................................................................
..|................................................................
.3o................................................................
....................................... - _Peter Bala_, Jul 08 2012

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 258, #10, F(n,k).

M. Drmota, Recursive Trees.

Diagonals give A001710, A006595. A109822, A130534.

t[n_, k_] := (n - 2)*t[n - 1, k] + t[n - 1, k - 1]; t[n_, n_] := 1; t[n_, 1] = n!/2; Table[t[n, k], {n, 10}, {k, n}] // Flatten (* Robert G. Wilson v, Jul 08 2012 *)

T(n, k) = (n-2)*T(n-1, k) + T(n-1, k-1), T(n, 1)=n!/2, T(n, n)=1.
From Peter Bala, Jul 08 2012: (Start)
E.g.f.: A(x,z) = x/(2-x){1/(1-z) - 1/(1-z)^(x-1)} = x*z + (x+x^2)*z^2/2! + (3*x+2*x^2+x^3)*z^3/3! + ....
A(x+1,z) is an e.g.f. for the row reverse of A109822 and A(x+2,z) generates the triangle of unsigned Stirling numbers of the first kind A130534 but with the first column omitted.
E.g.f. for column k: 1/(2^k*(1-x)) + (x-1)*sum {j = 0..k-1} 1/(j!*2^(k-j))*log^j(1/(1-x)) - see Drmota.
The row polynomial R(n,x) satisfies the recurrence equation R(n,x) = (x+n-2)*R(n-1,x) + (n-1)!*x with R(1,x) = x and also the recurrence R(n,x) = (x+2*n-3)*R(n-1,x) - (n-1)*(x+n-3)*R(n-2,x) with R(1,x) = x and R(2,x) = x+x^2. R(n,x) = {(x-1)*x*(x+1)*...*(x+n-2) - n!}/(x-2).
(End)

More terms from Larry Reeves (larryr(AT)acm.org), Jan 31 2001

A126682 Square pyramid giving coefficients of Carlo Wood's polynomials, read by successive slices, each slice being read row by row.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 3, 2, 2, 9, 7, 1, 6, 11, 1, 6, 11, 6, 3, 22, 45, 26, 3, 26, 69, 46, 1, 10, 35, 50, 1, 10, 35, 50, 24, 4, 45, 170, 255, 126, 6, 75, 320, 525, 274, 4, 55, 270, 545, 326, 1, 15, 85, 225, 274, 1, 15, 85, 225, 274, 120, 5, 81, 485, 1335, 1670, 744, 10

Offset: 1

N. J. A. Sloane, Feb 14 2007

There is no standard method for converting a pyramid of numbers to a sequence. This seems as good a solution as any.
See the link for further information and more terms.
The first row of each slice seems to coincide with the first row of each slice of A335442. That row from the n-th slice seems to be the coefficients of the polynomial (x+1) * ... * (x+n-1), i.e., the reversed row n-1 of A130534. - Andrey Zabolotskiy, Jun 26 2022

			Slice 1:
  1
Slice 2:
  1 1
  1 3
Slice 3:
  1  3  2
  2  9  7
  1  6 11
Slice 4:
  1  6 11  6
  3 22 45 26
  3 26 69 46
  1 10 35 50
Note that in Part 4 of the linked file, the order of the rows is reversed, while in its Part 1 the order of both rows and columns is reversed.

N. J. A. Sloane, Notes on Carlo Wood's Polynomials

Cf. A126671, A130534, A335442.

The sole term 1 of slice 1 inserted by Andrey Zabolotskiy, Jun 26 2022

A228094 Triangle starting at row 3 read by rows of the number of permutations in the n-th Dihedral group which are the product of k disjoint cycles, d(n,k), n >= 3, 1 <= k <= n.

Original entry on oeis.org

2, 3, 1, 2, 3, 2, 1, 4, 0, 5, 0, 1, 2, 2, 4, 3, 0, 1, 6, 0, 0, 7, 0, 0, 1, 4, 2, 0, 5, 4, 0, 0, 1, 6, 0, 2, 0, 9, 0, 0, 0, 1, 4, 4, 0, 0, 6, 5, 0, 0, 0, 1, 10, 0, 0, 0, 0, 11, 0, 0, 0, 0, 1, 4, 2, 2, 2, 0, 7, 6, 0, 0, 0, 0, 1, 12, 0, 0, 0, 0, 0, 13, 0, 0, 0, 0, 0, 1

Offset: 3

Robert A. Beeler, Aug 09 2013

The multivariable row polynomials give n times the cycle index for the Dihedral group D_n, called Z(D_n) (see the MathWorld link with the Harary reference). For example, 12*Z(D_6) = 2*(y_6)^1 + 2*(y_3)^2 + 4*(y_2)^3+3*(y_1)^2*(y_2)^2 + 1*(y_1)^6.

			Triangle begins
   2, 3, 1;
   2, 3, 2, 1;
   4, 0, 5, 0, 1;
   2, 2, 4, 3, 0,  1;
   6, 0, 0, 7, 0,  0, 1;
   4, 2, 0, 5, 4,  0, 0, 1;
   6, 0, 2, 0, 9,  0, 0, 0, 1;
   4, 4, 0, 0, 6,  5, 0, 0, 0, 1;
  10, 0, 0, 0, 0, 11, 0, 0, 0, 0, 1;
   4, 2, 2, 2, 0,  7, 6, 0, 0, 0, 0, 1;
   ...

Robert A. Beeler, How to Count: An Introduction to Combinatorics and Its Applications, Springer International Publishing, 2015. See Theorem 8.4.12 at pp. 246-247.
Frank Harary and Edgar M. Palmer, Graphical Enumeration, Academic Press, 1973, p. 37.

Stefano Spezia, First 150 rows of the triangle, flattened
Eric Weisstein's World of Mathematics, Cycle Index.

Cf. A000010, A054523, A130534.

d[n_,k_]:=If[Divisible[n,k],EulerPhi[n/k],0]+If[OddQ[n]&&k==(n+1)/2,n,If[EvenQ[n]&&(k==n/2||k==(n+2)/2),n/2,0]]; Table[d[n,k],{n,3,12},{k,n}]//Flatten (* Stefano Spezia, Jun 26 2023 *)

d(n,k) = A054523(n,k) + d'(n,k), where: If n is odd, then d'(n,k)= n when k=(n+1)/2 and d'(n,k)=0 otherwise. If n is even, then d'(n,k)=n/2 when k=n/2, (n+2)/2 and d'(n,k)=0 otherwise.

Terms corrected by Stefano Spezia, Jun 30 2023

A237996 Triangular array read by rows. T(n,k) is the number of even permutations of {1,2,...,n} that have exactly k cycles, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 11, 0, 1, 0, 24, 0, 35, 0, 1, 0, 0, 274, 0, 85, 0, 1, 0, 720, 0, 1624, 0, 175, 0, 1, 0, 0, 13068, 0, 6769, 0, 322, 0, 1, 0, 40320, 0, 118124, 0, 22449, 0, 546, 0, 1, 0, 0, 1026576, 0, 723680, 0, 63273, 0, 870, 0, 1

Offset: 0

Geoffrey Critzer, Feb 16 2014

Row sums = A001710(n).

If a permutation of {1, 2, ..., n} is written as a product of m disjoint cycles (where the fixed points of the permutation are viewed as 1-cycles) then the parity of the permutation is (-1)^(n-m). It is an even permutation if the number of cycles of even length is even (possibly zero), and it is an odd permutation if the number of cycles of even length is odd. - Peter Bala, Jun 25 2024

			Triangle begins:
  1;
  0,     1;
  0,     0,       1;
  0,     2,       0,      1;
  0,     0,      11,      0,      1;
  0,    24,       0,     35,      0,     1;
  0,     0,     274,      0,     85,     0,     1;
  0,   720,       0,   1624,      0,   175,     0,   1;
  0,     0,   13068,      0,   6769,     0,   322,   0,   1;
  0, 40320,       0, 118124,      0, 22449,     0, 546,   0, 1;
  0,     0, 1026576,      0, 723680,     0, 63273,   0, 870, 0, 1;
  ...

J. Riordan, Introduction to Combinatorial Analysis, Wiley, 1958, page 87, problem # 20.

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Parity of a permutaion

Cf. A008275, A130534, A132393, A164652.

with(combinat):
b:= proc(n, i, t) option remember; expand(`if`(n=0, t, `if`(i<1,
      0, add(x^j*multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j,
      i-1, irem(t+`if`(irem(i,2)=0, j, 0), 2)), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 1)):
seq(T(n), n=0..12);  # Alois P. Heinz, Mar 09 2015
# Alternative:
A132393 := (n, k) -> abs(Stirling1(n, k)):
T := (n, k) -> ifelse((n::even and k::even) or (n::odd and k::odd), A132393(n, k),
0): seq(seq(T(n, k), k = 0..n), n = 0..9);  # Peter Luschny, Jun 26 2024

nn=11;a=Log[((1+x)/(1-x))^(1/2)];b=Log[1/(1-x^2)^(1/2)];Table[Take[(Range[0,nn]!CoefficientList[Series[Exp[y a]Cosh[y b] ,{x,0,nn}],{x,y}])[[n]],n],{n,1,nn}]//Grid

E.g.f.: exp(y*A(x))*cosh(y*B(x)) where A(x) = log(((1 + x)/(1 - x))^(1/2)) and B(x) = log((1/(1-x^2))^(1/2)).
From Peter Bala, Jun 25 2024: (Start)
If n and k are both even or both odd, then T(n, k) is equal to the Stirling cycle number |s(n, k)| = A132393(n, k), and 0 otherwise.
n-th row polynomial R(n, x) = (1/2)*( x*(x + 1)*...*(x + n + 1) + x*(x - 1)*...*(x - n - 1) ).
For n >= 1, the zeros of R(n, x) are purely imaginary. (End)

A292222 Triangle corresponding to the partition array of the M_1 multinomials (A036038).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 10, 12, 24, 1, 15, 50, 60, 120, 1, 41, 180, 300, 360, 720, 1, 63, 497, 1260, 2100, 2520, 5040, 1, 162, 1484, 6496, 10080, 16800, 20160, 40320, 1, 255, 5154, 20916, 58464, 90720, 151200, 181440, 362880, 1, 637, 13680, 95640, 322560, 584640, 907200, 1512000, 1814400, 3628800

Offset: 1

Wolfdieter Lang, Sep 29 2017

Abramowitz-Stegun (A-St) M_1 multinomials as partition array (partitions in A-St order) are given in A036038. See this for details.
This is the sub-triangle of A226874(n,k) for n >= k >= 1 (here k=m).
The M_1 multinomials for a partition written in exponent form P = [1^e[1], 2^e[2], ... n^e[n]] with nonnegative e[j], for j =1, ..., n, is M_1(P) = n!/Product_{j=1..n} j!^e[j]. See the A-St link.

			The triangle T(n, m) begins:
n\m  1   2     3     4      5      6      7       8       9      10 ...
1:   1
2:   1   2
3:   1   3     6
4:   1  10    12    24
5:   1  15    50    60    120
6:   1  41   180   300    360    720
7:   1  63   497  1260   2100   2520   5040
8:   1 162  1484  6496  10080  16800  20160   40320
9:   1 255  5154 20916  58464  90720 151200  181440  362880
10:  1 637 13680 95640 322560 584640 907200 1512000 1814400 3628800
...
T(5, 3) =50 because the partitions are [1^2, 3^1] and [1^1, 2^2] with M_1 numbers 20 = A036038(5, 4) and 30 = A036038(5, 5), respectively, adding to 50.

Milton Abramowitz and Irene A. Stegun, editors, Multinomials: M_1, M_2 and M_3, Handbook of Mathematical Functions, December 1972, pp. 831-2.

Cf. A036038, A130534 (M_2 triangle = |Stirling1|), A008277 (M_3 triangle = Stirling2), A226874 (M_1 triangle including empty partition).

b[n_, i_, t_] := b[n, i, t] = If[t == 1, 1/n!, Sum[b[n - j, j, t - 1]/j!, {j, i, n/t}]];
t[n_, k_] := If[n*k == 0, If[n == k, 1, 0], n!*b[n, 1, k]];
Table[Table[t[n, k], {k, 1, n}], {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 29 2017, after Alois P. Heinz *)

T(n, m) = sum over the A036038 entries in row n with parts number m, for m >= n >= 1.

cp's OEIS Frontend

A260196 1, -3, followed by -1's.

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A358622 Regular triangle read by rows. T(n, k) = [[n, k]], where [[n, k]] are the second order Stirling cycle numbers (or second order reciprocal Stirling numbers). T(n, k) for 0 <= k <= n.

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A370518 Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(n,i) where TC(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+3,i+1)*(-1)^(i) for n >= 0, 0 <= k <= n.

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A387152 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} binomial(k, j)*|Stirling1(n, j)|.

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A387205 a(n) = (n - 1)!*(2 + Harmonic(n - 1)) if n >= 1, and a(0) = 1.

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A059418 Triangle T(n,k) arising from enumeration of permutations with ordered orbits, read by rows (1<=k<=n).

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A126682 Square pyramid giving coefficients of Carlo Wood's polynomials, read by successive slices, each slice being read row by row.

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A228094 Triangle starting at row 3 read by rows of the number of permutations in the n-th Dihedral group which are the product of k disjoint cycles, d(n,k), n >= 3, 1 <= k <= n.

A370518 Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n} binomial(n,i)(n-i)!Stirling1(i,k)TC(n,i) where TC(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)Stirling2(i+3,i+1)*(-1)^(i) for n >= 0, 0 <= k <= n.