A008517 -id:A008517 - cp's OEIS frontend

A336633 Triangle read by rows: T(n,k) is the number of generalized permutations related to the degenerate Eulerian numbers with exactly k ascents (0 <= k <= max(0,n-1)).

1, 1, 2, 2, 6, 16, 6, 24, 116, 116, 24, 120, 888, 1624, 888, 120, 720, 7416, 20984, 20984, 7416, 720, 5040, 67968, 270432, 419680, 270432, 67968, 5040, 40320, 682272, 3587904, 7861664, 7861664, 3587904, 682272, 40320, 362880, 7467840, 49701024, 144570624, 204403264, 144570624

Offset: 0

Orli Herscovici, Jul 28 2020

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..max(0,n-1)) begins:
     1;
     1;
     2,     2;
     6,    16,      6;
    24,   116,    116,     24;
   120,   888,   1624,    888,    120;
   720,  7416,  20984,  20984,   7416,   720;
  5040, 67968, 270432, 419680, 270432, 67968, 5040;
  ...

Orli Herscovici, Generalized permutations related to the degenerate Eulerian numbers, arXiv preprint arXiv:2007.13205 [math.CO], 2020.

Columns k = 0..1 give: A000142, A288964. Row sums give A007559.

Cf. A008292, A008517.

Tnk[0, 0] := 1; for n to N do
    for k from 0 to n do if 0 < k and k < n then Tnk[n, k] := (n + k)*Tnk[n - 1, k] + (2*n - k - 1)*Tnk[n - 1, k - 1]; else if k = 0 then Tnk[n, k] := (n + k)*Tnk[n - 1, k]; else Tnk[n, k] := 0; end if; end if; end do;
end do

T(n,k) = (n+k)*T(n-1,k) + (2*n-k-1)*T(n-1,k-1) for positive integers n and 0 <= k < n; T(0,0)=1 (or T(1,0)=1); otherwise T(n,k)=0.
From Peter Bala, Jan 08 2021: (Start)
The following remarks are all conjectures:
The e.g.f. (without the initial 1) A(x,t) = x + (2 + 2*t)*x^2/2! + (6 + 16*t + 6*t^2)*x^3/3! + ... satisfies the autonomous differential equation dA/dx = (1 + A)^2*(1 + t*A)^2.
The series reversion of A(x,t) with respect to x equals Integral_{u = 0..x} 1/((1 + u)^2*(1 + t*u)^2) du.
Let f(x,t) = (1 + x)^2*(1 + t*x)^2 and let D be the operator f(x,t)*d/dx. Then the (n+1)-th row polynomial = D^n(f(x,t)) evaluated at x = 0. (End)

A341111 T(n, k) = [x^k] M(n)Sum_{k=0..n} E2(n, k)binomial(-x + n - k, 2n), where E2 are the second-order Eulerian numbers A340556 and M(n) are the Minkowski numbers A053657. Triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= 2n+1.

Original entry on oeis.org

1, 0, 1, 1, 0, 10, 21, 14, 3, 0, 36, 96, 97, 47, 11, 1, 0, 12048, 36740, 45420, 29855, 11352, 2510, 300, 15, 0, 91200, 304480, 427348, 334620, 162255, 50787, 10302, 1310, 95, 3, 0, 109941120, 392583744, 603023624, 531477324, 300731214, 115291701, 30675678, 5682033, 719866, 59535, 2898, 63

Offset: 0

Peter Luschny, Feb 05 2021

			Triangle starts:
[0] 1;
[1] 0, 1,     1;
[2] 0, 10,    21,     14,     3;
[3] 0, 36,    96,     97,     47,     11,     1;
[4] 0, 12048, 36740,  45420,  29855,  11352,  2510,  300,   15;
[5] 0, 91200, 304480, 427348, 334620, 162255, 50787, 10302, 1310, 95, 3.

Cf. A053657, A163972, A008517, A201637, A340556, A341110 (row sums), A340556.

E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1)):
CoeffList := p -> [op(PolynomialTools:-CoefficientList(p, x))]:
mser := series((y/(exp(y)-1))^x, y, 29): m := n -> denom(coeff(mser, y, n)):
poly := n -> expand(m(n)*add(E2(n, k)*binomial(-x+n-k, 2*n), k = 0..n)):
for n from 0 to 6 do CoeffList(poly(n)) od;

M(n) = prod(i=1, #factor(n!)~, prime(i)^sum(k=0, #binary(n), floor((n-1)/((prime(i)-1)*prime(i)^k)))) \\ from A053657
rows_upto(n) = my(v1, v2); v1 = vector(n, i, 0); v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, v1[i] = (i+x)*(i+x-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+x)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 = vector(n+1, i, M(i)*Vecrev(v2[i])) \\ Mikhail Kurkov, Aug 27 2025

A102147 Second Eulerian transform of 1, 2, 3, 4, 5, ... (A000027).

Original entry on oeis.org

1, 1, 5, 35, 315, 3465, 45045, 675675, 11486475, 218243025, 4583103525, 105411381075, 2635284526875, 71152682225625, 2063427784543125, 63966261320836875, 2110886623587616875, 73881031825566590625

Offset: 1

Ross La Haye, Feb 14 2005

nonn

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 256.

Cf. A001147, A008517.

Apparently equals A051577(n-2), n > 1.

a(n) = Sum_{k=0..n} E(n,k) A000027(k), where E(n,k) is a second-order Eulerian number (A008517).

A156184 A generalized recursion triangle sequence : m=1; t(n,k)=(k + m - 1)t(n - 1, k, m) + (mn - k + 1 - m)*t(n - 1, k - 1, m).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 7, 16, 7, 1, 1, 11, 53, 53, 11, 1, 1, 16, 150, 318, 150, 16, 1, 1, 22, 380, 1554, 1554, 380, 22, 1, 1, 29, 892, 6562, 12432, 6562, 892, 29, 1, 1, 37, 1987, 25038, 82538, 82538, 25038, 1987, 37, 1, 1, 46, 4270, 89023, 480380, 825380, 480380

Offset: 0

Roger L. Bagula, Feb 05 2009

Row sums are: A054091;
{1, 2, 4, 10, 32, 130, 652, 3914, 27400, 219202, 1972820, ...}.
The sequence comes from a generalization of the recurrence for A008517.

			{1},
{1, 1},
{1, 2, 1},
{1, 4, 4, 1},
{1, 7, 16, 7, 1},
{1, 11, 53, 53, 11, 1},
{1, 16, 150, 318, 150, 16, 1},
{1, 22, 380, 1554, 1554, 380, 22, 1},
{1, 29, 892, 6562, 12432, 6562, 892, 29, 1},
{1, 37, 1987, 25038, 82538, 82538, 25038, 1987, 37, 1},
{1, 46, 4270, 89023, 480380, 825380, 480380, 89023, 4270, 46, 1}

Eric Weisstein's World of Mathematics, Second-Order Eulerian Triangle.

Cf. A054091, A054090, A008517, A156141.

m = 1; e[n_, 0, m_] := 1;
e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n;
e[n_, k_, m_] := (k + m - 1)e[n - 1, k, m] + (m*n - k + 1 - m)e[n - 1, k - 1, m];
Table[Table[e[n, k, m], {k, 0, n}], {n, 0, 10}];
Flatten[%]

t(n,k) = (k + m - 1)*t(n - 1, k, m) + (m*n - k + 1 - m)*t(n - 1, k - 1, m).

A156186 Triangle: m=3; e(n,k,n) = (k + m - 1)e(n - 1, k, m) + (mn - k + 1 - m)*e(n - 1, k - 1, m); t(n,k) = e(n,k,m) + e(n,n-k,m).

Original entry on oeis.org

2, 1, 1, 1, 6, 1, 1, 30, 30, 1, 1, 159, 360, 159, 1, 1, 1119, 3639, 3639, 1119, 1, 1, 10932, 41262, 57414, 41262, 10932, 1, 1, 136764, 582642, 898632, 898632, 582642, 136764, 1, 1, 2031933, 9957168, 16634718, 17182152, 16634718, 9957168, 2031933, 1, 1

Offset: 0

Roger L. Bagula, Feb 05 2009

			{2},
{1, 1},
{1, 6, 1},
{1, 30, 30, 1},
{1, 159, 360, 159, 1},
{1, 1119, 3639, 3639, 1119, 1},
{1, 10932, 41262, 57414, 41262, 10932, 1},
{1, 136764, 582642, 898632, 898632, 582642, 136764, 1},
{1, 2031933, 9957168, 16634718, 17182152, 16634718, 9957168, 2031933, 1},...

Cf. A054091, A054090, A008517, A156141.

m = 3; e[n_, 0, m_] := 1;
e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n;
e[n_, k_, m_] := (k + m - 1)e[n - 1, k, m] + (m*n - k + 1 - m)e[n - 1, k - 1, m];
Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
Table[Table[e[n, k, m] + e[n, n - k, m], {k, 0, n}], {n, 0, 10}];
Flatten[%]

m=3; e(n,k,n) = (k + m - 1)*e(n - 1, k, m) + (m*n - k + 1 - m)*e(n - 1, k - 1, m);

t(n,k) = e(n,k,m) + e(n,n-k,m).

A156188 Triangle: m=5; e(n,k,n)=(k + m - 1)e(n - 1, k, m) + (mn - k + 1 - m)*e(n - 1, k - 1, m); t(n,k)=e(n,k,m)+e(n,n-k,m).

Original entry on oeis.org

2, 1, 1, 1, 10, 1, 1, 80, 80, 1, 1, 775, 1520, 775, 1, 1, 10915, 25945, 25945, 10915, 1, 1, 213720, 542910, 624670, 542910, 213720, 1, 1, 5245530, 14690640, 16408670, 16408670, 14690640, 5245530, 1, 1, 151534685, 479956020, 553630850, 464654480

Offset: 0

Roger L. Bagula, Feb 05 2009

			{2},
{1, 1},
{1, 10, 1},
{1, 80, 80, 1},
{1, 775, 1520, 775, 1},
{1, 10915, 25945, 25945, 10915, 1},
{1, 213720, 542910, 624670, 542910, 213720, 1},
{1, 5245530, 14690640, 16408670, 16408670, 14690640, 5245530, 1},...

Cf. A054091, A054090, A008517, A156141.

m = 5; e[n_, 0, m_] := 1;
e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n;
e[n_, k_, m_] := (k + m - 1)e[n - 1, k, m] + (m*n - k + 1 - m)e[n - 1, k - 1, m];
Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
Table[Table[e[n, k, m] + e[n, n - k, m], {k, 0, n}], {n, 0, 10}];
Flatten[%]

m=5; e(n,k,n) = (k + m - 1)*e(n - 1, k, m) + (m*n - k + 1 - m)*e(n - 1, k - 1, m);

t(n,k) = e(n,k,m) + e(n,n-k,m).

A156278 A higher order recursion triangle sequence: m=3;l=3;e(n,k,m)=(lk + m - 1)e(n - 1, k, m) + (mn - l*k + 1 - m)e(n - 1, k - 1, m).

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 52, 44, 1, 1, 270, 716, 187, 1, 1, 1363, 8428, 7069, 762, 1, 1, 6831, 85143, 162039, 60151, 3065, 1, 1, 34174, 790440, 2889288, 2462504, 473162, 12280, 1, 1, 170892, 6972826, 44429208, 72035800, 32668794, 3557734, 49143, 1, 1, 854485

Offset: 0

Roger L. Bagula and Gary W. Adamson, Feb 07 2009

Row sums are:
{2, 2, 4, 22, 196, 2350, 35248, 634462, 13323700, 319768798, 8633757544,...}.
The MacMahon level generalization that I was looking for:
I can get the Sierpinski Pascal mostly by this method too.
I did it by looking at the three variables {n,k,m} as being a 3d plane and the General -Sierpinski-Pascal like
{{m,0,0},
{0,-m,0},
{0,0,1}}. {n,k,1}
and the General Eulerian as being like:
{{m,0,0},
{0,-1,1},
{0,0-m}}. {n,k,1}
So the MacMahon is the next quantum step up in k:
{{m,0,0},
{0,-2,1},
{0,0-m}}. {n,k,1}
The further generalization adds a new quantum variable l:
{{m,0,0},
{0,-l,1},
{0,0-m}}. {n,k,1}
This recursive result seems to give a much more general type of combinatorial triangle sequence.

			{1},
{1, 1},
{1, 9, 1},
{1, 52, 44, 1},
{1, 270, 716, 187, 1},
{1, 1363, 8428, 7069, 762, 1},
{1, 6831, 85143, 162039, 60151, 3065, 1},
{1, 34174, 790440, 2889288, 2462504, 473162, 12280, 1},
{1, 170892, 6972826, 44429208, 72035800, 32668794, 3557734, 49143, 1},
{1, 854485, 59542232, 621204982, 1719368528, 1491834898, 397842620, 26034427, 196598, 1}

A008517

m = 3; l = 3;
e[n_, 0, m_] := 1; e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n
e[n_, k_, m_] := (l*k + m - 1)e[ n - 1, k, m] + (m*n - l*k + 1 - m)e[n - 1, k - 1, m];
Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
Flatten[%]

m=3;l=3;

e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).

A156280 A higher order recursion triangle sequence: m=4;l=4;e(n,k,m)=(lk + m - 1)e(n - 1, k, m) + (mn - l*k + 1 - m)e(n - 1, k - 1, m).

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 93, 71, 1, 1, 664, 1618, 370, 1, 1, 4665, 26430, 20112, 1869, 1, 1, 32676, 370035, 645270, 216519, 9368, 1, 1, 228757, 4756581, 15969645, 12502371, 2164135, 46867, 1, 1, 1601328, 58041316, 339432876, 509029014, 212305928

Offset: 0

Roger L. Bagula and Gary W. Adamson, Feb 07 2009

Row sums are:
{1, 1, 2, 14, 166, 2654, 53078, 1273870, 35668358, 1141387454, 41089948342,...}.
The MacMahon level generalization that I was looking for:
I can get the Sierpinski Pascal mostly by this method too.
I did it by looking at the three variables {n,k,m} as being a 3d plane and the General -Sierpinski-Pascal like
{{m,0,0},
{0,-m,0},
{0,0,1}}. {n,k,1}
and the General Eulerian as being like:
{{m,0,0},
{0,-1,1},
{0,0-m}}. {n,k,1}
So the MacMahon is the next quantum step up in k:
{{m,0,0},
{0,-2,1},
{0,0-m}}. {n,k,1}
The further generalization adds a new quantum variable l:
{{m,0,0},
{0,-l,1},
{0,0-m}}. {n,k,1}
This recursive result seems to give a much more general type of combinatorial triangle sequence.

			{1},
{1, 1},
{1, 12, 1},
{1, 93, 71, 1},
{1, 664, 1618, 370, 1},
{1, 4665, 26430, 20112, 1869, 1},
{1, 32676, 370035, 645270, 216519, 9368, 1},
{1, 228757, 4756581, 15969645, 12502371, 2164135, 46867, 1},
{1, 1601328, 58041316, 339432876, 509029014, 212305928, 20742624, 234366, 1},
{1, 11209329, 684892988, 6542526040, 16799641662, 13536529582, 3320027912, 193948962, 1171865, 1}

A008517

m = 4; l = 4;
e[n_, 0, m_] := 1; e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n
e[n_, k_, m_] := (l*k + m - 1)e[ n - 1, k, m] + (m*n - l*k + 1 - m)e[n - 1, k - 1, m];
Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
Flatten[%]

m=4;l=4;

e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).

A219512 Triangle of third-order Eulerian numbers: 3-Stirling permutations enumerated by ascents.

Original entry on oeis.org

1, 1, 3, 1, 12, 15, 1, 33, 141, 105, 1, 78, 786, 1830, 945, 1, 171, 3450, 17538, 26685, 10395, 1, 360, 13257, 125352, 396495, 435960, 135135, 1, 741, 46971, 753291, 4238811, 9356175, 7921305, 2027025, 1, 1506, 157956, 4046526, 37013166, 140913270, 233216460, 158799690, 34459425

Offset: 1

Peter Bala, Dec 11 2012

See A008292 for the triangle of Eulerian numbers and A008517 for the triangle of second-order Eulerian numbers (2-Stirling permutations).

A 3-Stirling permutation of order n is a permutation of the multiset {1,1,1,2,2,2,...,n,n,n} such that for each i, 1 <= i <= n, the elements occurring between two occurrences of i are at least i. This triangle enumerates 3-Stirling permutations by ascents. Examples are given below.

			Triangle begins
.n\k.|..0....1......2.......3......4........5.......6
= = = = = = = = = = = = = = = = = = = = = = = = = = =
..1..|..1
..2..|..1....3
..3..|..1...12.....15
..4..|..1...33....141.....105
..5..|..1...78....786....1830....945
..6..|..1..171...3450...17538..26685....10395
..7..|..1..360..13257..125352.396495...435960..135135
...
Example of recurrence: T(5,2) = 3*141 + 11*33 = 786.
Row 2 = [1,3]. The 3-Stirling permutations of order 2 are obtained by inserting the string 222 into one of the four available positions in the string 111, giving 222111, 122211, 112221 and 111222. The first permutation has no ascents while the remaining three permutations each have 1 ascent.

J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013.
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. II. Applications, arXiv preprint arXiv:1307.5624 [math.CO], 2013.
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Generalized Stirling permutations and forests: Higher-order Eulerian and Ward numbers, Electronic Journal of Combinatorics 22(3) (2015), #P3.37.
Tian-Xiao He, The mth-order Eulerian Numbers, arXiv:2312.17153 [math.CO], 2023.
Svante Janson, Markus Kuba, and Alois Panholzer, Generalized Stirling permutations, families of increasing trees and urn models arXiv:0805.4084v1 [math.CO], 2008.
SeungKyung Park, Inverse descents of r-multipermutations, Discrete Mathematics 132, 1-3, 215-229, 1994.
Grzegorz Rzadkowski and Malgorzata Urlinska, A Generalization of the Eulerian Numbers, arXiv preprint arXiv:1612.06635 [math.CO], 2016-2017.
Umesh Shankar, Log-concavity of rows of triangular arrays satisfying a certain super-recurrence, arXiv:2508.12467 [math.CO], 2025. See p. 4.

Cf. A001147 (main diagonal), A007559 (row sums), A008292, A008517.

T[n_, k_] /; 1 <= k <= n-1 := T[n, k] = (k+1) T[n-1, k] + (3n-k-2) T[n-1, k-1]; T[, 0] = 1; T[, _] = 0;
Table[T[n, k], {n, 1, 9}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 12 2019 *)

Recurrence equation: T(n+1,k) = (k + 1)*T(n,k) + (3*n - k + 1)*T(n,k-1).
Let B(x,t) = Integral_{x' = 0..x} 1/((1 + x')*( 1+ t*x')^3) dx' = x - (1 + 3*t)*x^2/2 + (1 + 3*t + 6*t^2)*x^3/3 - (1 + 3*t + 6*t^2 + 10*t^3)*x^4/4 + ....
The e.g.f. A(x,t) appears to be the series reversion (with respect to x) B(x,t)^<-1> = x + (1 + 3*t)*x^2/2! + (1 + 12*t + 15*t^2)*x^3/3! + .... If true, then the e.g.f. A(x,t) satisfies the autonomous differential equation dA/dx = (1 + A)*(1 + t*A)^3, A(0,t) = 0. Also if we let D be the operator (1 + x)*(1 + t*x)^3*d/dx then the (n+1)-th row polynomial R(n+1,t) = D^n(x) evaluated at x = 0.

A321591 Partitioned 2nd-order Eulerian numbers forming an "Eulerian pyramid" (tetrahedron).

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 4, 1, 4, 1, 1, 11, 11, 11, 36, 11, 1, 11, 11, 1, 1, 26, 26, 66, 196, 66, 26, 196, 196, 26, 1, 26, 66, 26, 1, 1, 57, 57, 302, 848, 302, 302, 1898, 1898, 302, 57, 848, 1898, 848, 57, 1, 57, 302, 302, 57, 1, 1, 120, 120, 1191, 3228, 1191, 2416, 13644

Offset: 0

Gregory Gerard Wojnar, Nov 13 2018

For N+1 = i+j+k, let P(N+1;i,j,k) = (N+1-i)*P(N;i-1,j,k) + (N+1-j)*P(N;i,j-1,k) + (N+1-k)*P(N;i,j,k-1), with P(N;i,j,k) invariant upon permutation of the indices i,j,k, also P(N;N,0,0)=1 and P(N;i,j,k) = 0 if i or j or k is negative. The indexing of these values is shown explicitly in the examples.

The row sums are the second-order Eulerian numbers, A008517; precisely, Sum_{(j,k)|j+k=N-i} P(N;i,j,k) = <> = T(N+1,i+1) of A008517. The row sum of row i=N of slice N is (N+1)!. The sum of all entries in slice N is (2*N+1)!!. The edges of the N-th triangular slice of the pyramid are row (N+1) of the first-order Eulerian triangle, A008292.

			The first few slices of the tetrahedron (and row sums) are:
  1                  (1); i=0, N=0, (j,k)=(0,0)
------------------------
   1                 (1); i=0, N=1, (j,k)=(0,0)
  1 1                (2); i=1, N=1, (j,k)=(1,0) (0,1)
------------------------
    1                (1); i=0, N=2, (j,k)=(0,0)
   4 4               (8); i=1, N=2, (j,k)=(1,0) (0,1)
  1 4 1              (6); i=2, N=2, (j,k)=(2,0) (1,1) (0,2)
------------------------
      1              (1); i=0, N=3, (j,k)=(0,0)
    11 11           (22); i=1, N=3, (j,k)=(1,0) (0,1)
   11 36 11         (58); i=2, N=3, (j,k)=(2,0) (1,1) (0,2)
  1 11 11  1        (24); i=3, N=3, (j,k)=(3,0) (2,1) (1,2) (0,3)
------------------------
         1           (1); i=0, N=4, (j,k)=(0,0)
       26 26        (52); i=1, N=4, (j,k)=(1,0) (0,1)
     66 196 66     (328); i=2, N=4, (j,k)=(2,0) (1,1) (0,2)
   26 196 196 26   (444); i=3, N=4, (j,k)=(3,0) (2,1) (1,2) (0,3)
  1  26  66  26 1  (120); i=4, N=4, (j,k)=(4,0) (3,1) (2,2) (1,3) (0,4)

cp's OEIS Frontend

A336633 Triangle read by rows: T(n,k) is the number of generalized permutations related to the degenerate Eulerian numbers with exactly k ascents (0 <= k <= max(0,n-1)).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

A341111 T(n, k) = [x^k] M(n)*Sum_{k=0..n} E2(n, k)*binomial(-x + n - k, 2*n), where E2 are the second-order Eulerian numbers A340556 and M(n) are the Minkowski numbers A053657. Triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= 2*n+1.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

PARI

A102147 Second Eulerian transform of 1, 2, 3, 4, 5, ... (A000027).

Views

Author

Keywords

References

Crossrefs

Formula

A156184 A generalized recursion triangle sequence : m=1; t(n,k)=(k + m - 1)*t(n - 1, k, m) + (m*n - k + 1 - m)*t(n - 1, k - 1, m).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A156186 Triangle: m=3; e(n,k,n) = (k + m - 1)*e(n - 1, k, m) + (m*n - k + 1 - m)*e(n - 1, k - 1, m); t(n,k) = e(n,k,m) + e(n,n-k,m).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A156188 Triangle: m=5; e(n,k,n)=(k + m - 1)*e(n - 1, k, m) + (m*n - k + 1 - m)*e(n - 1, k - 1, m); t(n,k)=e(n,k,m)+e(n,n-k,m).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A156278 A higher order recursion triangle sequence: m=3;l=3;e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A156280 A higher order recursion triangle sequence: m=4;l=4;e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A219512 Triangle of third-order Eulerian numbers: 3-Stirling permutations enumerated by ascents.

Views

Author

Keywords

A341111 T(n, k) = [x^k] M(n)Sum_{k=0..n} E2(n, k)binomial(-x + n - k, 2n), where E2 are the second-order Eulerian numbers A340556 and M(n) are the Minkowski numbers A053657. Triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= 2n+1.

A156184 A generalized recursion triangle sequence : m=1; t(n,k)=(k + m - 1)t(n - 1, k, m) + (mn - k + 1 - m)*t(n - 1, k - 1, m).

A156186 Triangle: m=3; e(n,k,n) = (k + m - 1)e(n - 1, k, m) + (mn - k + 1 - m)*e(n - 1, k - 1, m); t(n,k) = e(n,k,m) + e(n,n-k,m).

A156188 Triangle: m=5; e(n,k,n)=(k + m - 1)e(n - 1, k, m) + (mn - k + 1 - m)*e(n - 1, k - 1, m); t(n,k)=e(n,k,m)+e(n,n-k,m).

A156278 A higher order recursion triangle sequence: m=3;l=3;e(n,k,m)=(lk + m - 1)e(n - 1, k, m) + (mn - l*k + 1 - m)e(n - 1, k - 1, m).

A156280 A higher order recursion triangle sequence: m=4;l=4;e(n,k,m)=(lk + m - 1)e(n - 1, k, m) + (mn - l*k + 1 - m)e(n - 1, k - 1, m).