A211212 -id:A211212 - cp's OEIS frontend

A278074 Triangle read by rows, coefficients of the polynomials P(m, n) = Sum_{k=1..n} binomial(mn, mk)* P(m, n-k)*z with P(m, 0) = 1 and m = 4.

1, 0, 1, 0, 1, 70, 0, 1, 990, 34650, 0, 1, 16510, 2702700, 63063000, 0, 1, 261630, 213519150, 17459442000, 305540235000, 0, 1, 4196350, 17651304000, 4350310965000, 231905038365000, 3246670537110000

Offset: 0

Peter Luschny, Jan 22 2017

			Triangle starts:
[1]
[0, 1]
[0, 1,     70]
[0, 1,    990,     34650]
[0, 1,  16510,   2702700,    63063000]
[0, 1, 261630, 213519150, 17459442000, 305540235000]

Robert Israel, Table of n, a(n) for n = 0..8510

Cf. A014608 (diagonal), A243665 (row sums), A211212 (alternating row sums), A281480 (central coefficients).
Cf. A097805 (m=0), A131689 (m=1), A241171 (m=2), A278073 (m=3).
Cf. A327024 (refinement).

P := proc(m,n) option remember; if n = 0 then 1 else
add(binomial(m*n,m*k)* P(m,n-k)*x, k=1..n) fi end:
for n from 0 to 6 do PolynomialTools:-CoefficientList(P(4,n), x) od;
# Alternatively:
A278074_row := proc(n) 1/(1-t*((cosh(x)+cos(x))/2-1)); expand(series(%,x,4*n+1));
(4*n)!*coeff(%,x,4*n); PolynomialTools:-CoefficientList(%,t) end:
for n from 0 to 5 do A278074_row(n) od;

With[{m = 4}, Table[Expand[j!*SeriesCoefficient[1/(1 - t*(MittagLefflerE[m, x^m] - 1)), {x, 0, j}]], {j, 0, 24, m}]];
Function[arg, CoefficientList[arg, t]] /@ % // Flatten

# uses [P from A278073]
def A278074_row(n): return list(P(4, n))
for n in (0..6): print(A278074_row(n)) # Peter Luschny, Mar 24 2020

E.g.f.: 1/(1-t*((cosh(x)+cos(x))/2-1)), nonzero terms.

A181985 Generalized Euler numbers. Square array A(n,k), n >= 1, k >= 0, read by antidiagonals. A(n,k) = n-alternating permutations of length n*k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 19, 61, 1, 1, 1, 69, 1513, 1385, 1, 1, 1, 251, 33661, 315523, 50521, 1, 1, 1, 923, 750751, 60376809, 136085041, 2702765, 1, 1, 1, 3431, 17116009, 11593285251, 288294050521, 105261234643, 199360981, 1

Offset: 1

Peter Luschny, Apr 04 2012

For an integer n > 0, a permutation s = s_1...s_k is an n-alternating permutation if it has the property that s_i < s_{i+1} if and only if n divides i.
The classical Euler numbers count 2-alternating permutations of length 2n.
Ludwig Seidel gave in 1877 an efficient algorithm to compute the coefficients of sec which carries immediately over to the computation of the generalized Euler numbers (see the Maple script).

			n\k [0][1]  [2]       [3]            [4]                 [5]
[1]  1, 1,   1,        1,             1,                   1
[2]  1, 1,   5,       61,          1385,               50521  [A000364]
[3]  1, 1,  19,     1513,        315523,           136085041  [A002115]
[4]  1, 1,  69,    33661,      60376809,        288294050521  [A211212]
[5]  1, 1, 251,   750751,   11593285251,     613498040952501
[6]  1, 1, 923, 17116009, 2301250545971, 1364944703949044401
       [A030662][A211213]   [A181991]
The (n,n)-diagonal is A181992.

Peter Luschny, An old operation on sequences: the Seidel transform.
Ludwig Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187. [USA access only through the HATHI TRUST Digital Library]
Ludwig Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187. [Access through ZOBODAT]

Cf. A181937, A000364, A002115, A030662, A211212, A211213, A181991, A181992.

A181985_list := proc(n, len) local E, dim, i, k;
dim := n*(len-1); E := array(0..dim, 0..dim); E[0, 0] := 1;
for i from 1 to dim do
   if i mod n = 0 then E[i, 0] := 0 ;
      for k from i-1 by -1 to 0 do E[k, i-k] := E[k+1, i-k-1] + E[k, i-k-1] od;
   else E[0, i] := 0;
      for k from 1 by 1 to i do E[k, i-k] := E[k-1, i-k+1] + E[k-1, i-k] od;
   fi od;
seq(E[0, n*k], k=0..len-1) end:
for n from 1 to 6 do print(A181985_list(n, 6)) od;

nmax = 9; A181985[n_, len_] := Module[{e, dim = n*(len - 1)}, e[0, 0] = 1; For[i = 1, i <= dim, i++, If[Mod[i, n] == 0 , e[i, 0] = 0 ; For[k = i-1, k >= 0, k--, e[k, i-k] = e[k+1, i-k-1] + e[k, i-k-1] ], e[0, i] = 0; For[k = 1, k <= i, k++, e[k, i-k] = e[k-1, i-k+1] + e[k-1, i-k] ]; ]]; Table[e[0, n*k], { k, 0, len-1}]]; t = Table[A181985[n, nmax], {n, 1, nmax}]; a[n_, k_] := t[[n, k+1]]; Table[a[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jun 27 2013, translated and adapted from Maple *)

def A181985(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return (-1)^n*sum((-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
for m in (1..6): print([A181985(m, n) for n in (0..7)]) # Peter Luschny, Aug 10 2015

A292604 Triangle read by rows, coefficients of generalized Eulerian polynomials F_{2}(x).

Original entry on oeis.org

1, 1, 0, 5, 1, 0, 61, 28, 1, 0, 1385, 1011, 123, 1, 0, 50521, 50666, 11706, 506, 1, 0, 2702765, 3448901, 1212146, 118546, 2041, 1, 0, 199360981, 308869464, 147485535, 24226000, 1130235, 8184, 1, 0

Offset: 0

Peter Luschny, Sep 20 2017

The generalized Eulerian polynomials F_{m}(x) are defined F_{m; 0}(x) = 1 for all m >= 0 and for n > 0:
F_{0; n}(x) = Sum_{k=0..n} A097805(n, k)*(x-1)^(n-k) with coeffs. in A129186.
F_{1; n}(x) = Sum_{k=0..n} A131689(n, k)*(x-1)^(n-k) with coeffs. in A173018.
F_{2; n}(x) = Sum_{k=0..n} A241171(n, k)*(x-1)^(n-k) with coeffs. in A292604.
F_{3; n}(x) = Sum_{k=0..n} A278073(n, k)*(x-1)^(n-k) with coeffs. in A292605.
F_{4; n}(x) = Sum_{k=0..n} A278074(n, k)*(x-1)^(n-k) with coeffs. in A292606.
The case m = 1 are the Eulerian polynomials whose coefficients are the Eulerian numbers which are displayed in Euler's triangle A173018.
Evaluated at x in {-1, 1, 0} these families of polynomials give for the first few m:
F_{m} :   F_{0}    F_{1}    F_{2}    F_{3}    F_{4}
x = -1:  A165326  A155585  A002105  A292609  A292607
x =  1:  A000012  A000142  A000680  A014606  A014608  ... (m*n)!/m!^n
x =  0:    --     A000012  A000364  A002115  A211212  ... m-alternating permutations of length m*n.
Note that the constant terms of the polynomials are the generalized Euler numbers as defined in A181985. In this sense generalized Euler numbers are also generalized Eulerian numbers.

			Triangle starts:
[n\k][    0        1        2       3     4  5  6]
--------------------------------------------------
[0][      1]
[1][      1,       0]
[2][      5,       1,       0]
[3][     61,      28,       1,      0]
[4][   1385,    1011,     123,      1,    0]
[5][  50521,   50666,   11706,    506,    1, 0]
[6][2702765, 3448901, 1212146, 118546, 2041, 1, 0]

G. Frobenius. Über die Bernoullischen Zahlen und die Eulerschen Polynome. Sitzungsber. Preuss. Akad. Wiss. Berlin, pages 200-208, 1910.

F_{0} = A129186, F_{1} = A173018, F_{2} is this triangle, F_{3} = A292605, F_{4} = A292606.
First column: A000364. Row sums: A000680. Alternating row sums: A002105.
Cf. A181985, A241171.

Coeffs := f -> PolynomialTools:-CoefficientList(expand(f), x):
A292604_row := proc(n) if n = 0 then return [1] fi;
add(A241171(n, k)*(x-1)^(n-k), k=0..n); [op(Coeffs(%)), 0] end:
for n from 0 to 6 do A292604_row(n) od;

T[n_, k_] /; 1 <= k <= n := T[n, k] = k (2 k - 1) T[n - 1, k - 1] + k^2 T[n - 1, k]; T[, 1] = 1; T[, _] = 0;
F[2, 0][] = 1; F[2, n][x_] := Sum[T[n, k] (x - 1)^(n - k), {k, 0, n}];
row[n_] := If[n == 0, {1}, Append[CoefficientList[ F[2, n][x], x], 0]];
Table[row[n], {n, 0, 7}] (* Jean-François Alcover, Jul 06 2019 *)

def A292604_row(n):
    if n == 0: return [1]
    S = sum(A241171(n, k)*(x-1)^(n-k) for k in (0..n))
    return expand(S).list() + [0]
for n in (0..6): print(A292604_row(n))

F_{2; n}(x) = Sum_{k=0..n} A241171(n, k)*(x-1)^(n-k) for n>0 and F_{2; 0}(x) = 1.

A292606 Triangle read by rows, coefficients of generalized Eulerian polynomials F_{4;n}(x).

Original entry on oeis.org

1, 1, 0, 69, 1, 0, 33661, 988, 1, 0, 60376809, 2669683, 16507, 1, 0, 288294050521, 17033188586, 212734266, 261626, 1, 0, 3019098162602349, 223257353561605, 4297382231090, 17634518610, 4196345, 1, 0

Offset: 0

Peter Luschny, Sep 26 2017

See the comments in A292604.

			Triangle starts:
[n\k][          0            1          2       3   4   5]
--------------------------------------------------
[0] [           1]
[1] [           1,           0]
[2] [          69,           1,         0]
[3] [       33661,         988,         1,      0]
[4] [    60376809,     2669683,     16507,      1,  0]
[5] [288294050521, 17033188586, 212734266, 261626,  1,  0]

F_{0} = A129186, F_{1} = A173018, F_{2} = A292604, F_{3} = A292605, F_{4} is this triangle.
First column: A211212. Row sums: A014608. Alternating row sums: A292607.
Cf. A181985.

Coeffs := f -> PolynomialTools:-CoefficientList(expand(f), x):
A292606_row := proc(n) if n = 0 then return [1] fi;
add(A278074(n, k)*(x-1)^(n-k), k=0..n); [op(Coeffs(%)), 0] end:
for n from 0 to 6 do A292606_row(n) od;

# uses[A278074_row from A278074]
def A292606_row(n):
    if n == 0: return [1]
    L = A278074_row(n)
    S = sum(L[k]*(x-1)^(n-k) for k in (0..n))
    return expand(S).list() + [0]
for n in (0..5): print(A292606_row(n))

F_{4; n}(x) = Sum_{k=0..n} A278074(n, k)*(x-1)^(n-k) for n>0 and F_{4; 0}(x) = 1.

A327024 Ordered set partitions of the set {1, 2, ..., 4*n} with all block sizes divisible by 4, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

Original entry on oeis.org

1, 1, 1, 70, 1, 990, 34650, 1, 3640, 12870, 2702700, 63063000, 1, 9690, 251940, 26453700, 187065450, 17459442000, 305540235000, 1, 21252, 1470942, 2704156, 154448910, 8031343320, 9465511770, 374796021600, 3975514943400, 231905038365000, 3246670537110000

Offset: 0

Peter Luschny, Aug 27 2019

T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 4. For instance 4*P(4, .) = [[16], [12, 4], [8, 8], [8, 4, 4], [4, 4, 4, 4]].

			Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1;    70]
[3] [1;   990;   34650]
[4] [1;  3640,   12870;  2702700;  63063000]
[5] [1;  9690,  251940; 26453700, 187065450; 17459442000; 305540235000]
[6] [1; 21252, 1470942,  2704156; 154448910,  8031343320,   9465511770;
     374796021600, 3975514943400; 231905038365000; 3246670537110000]
.
T(4, 1) = 3640 because [12, 4] is the integer partition 4*P(4, 1) in the canonical order and there are 1820 set partitions which have the shape [12, 4]. Finally, since the order of the sets is taken into account, one gets 2!*1820 = 3640.

Row sums: A243665, alternating row sums: A211212, main diagonal: A014608, central column: A281480, by length: A278074.
Cf. A178803 (m=0), A133314 (m=1), A327022 (m=2), A327023 (m=3), this sequence (m=4).
Cf. A080577, A000041, A072233.

# uses[GenOrdSetPart from A327022]
def A327024row(n): return GenOrdSetPart(4, n)
for n in (0..6): print(A327024row(n))

A260877 Square array read by ascending antidiagonals: number of m-shape Euler numbers.

Original entry on oeis.org

1, 1, -1, 1, -1, 1, 1, -1, 1, -5, 1, -1, 5, -1, 21, 1, -1, 19, -61, 1, -105, 1, -1, 69, -1513, 1385, -1, 635, 1, -1, 251, -33661, 315523, -50521, 1, -4507, 1, -1, 923, -750751, 60376809, -136085041, 2702765, -1, 36457, 1, -1, 3431, -17116009, 11593285251

Offset: 1

Peter Luschny, Aug 09 2015

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n. It is ordered if the positions of the blocks are taken into account.
M-shape Euler numbers count the ordered m-shape set partitions which have even length minus the number of such partitions which have odd length.
If m=0 all possible sizes are zero. Thus m-shape Euler numbers count the ordered integer partitions of n into an even number of parts minus the number of ordered integer partitions of n into an odd number of parts (A260845).
If m=1 the set is {1,2,...,n} and the set of all possible sizes are the integer partitions of n. Thus the Euler numbers count the ordered set partitions which have even length minus the set partitions which have odd length (A033999).
If m=2 the set is {1,2,...,2n} and the 2-shape Euler numbers count the ordered set partitions with even blocks which have even length minus the number of partitions with even blocks which have odd length (A028296).

			[ n ] [0   1   2       3         4              5                 6]
[ m ] --------------------------------------------------------------
[ 0 ] [1, -1,  1,     -5,       21,          -105,              635] A260845
[ 1 ] [1, -1,  1,     -1,        1,            -1,                1] A033999
[ 2 ] [1, -1,  5,    -61,     1385,        -50521,          2702765] A028296
[ 3 ] [1, -1, 19,  -1513,   315523,    -136085041,     105261234643] A002115
[ 4 ] [1, -1, 69, -33661, 60376809, -288294050521, 3019098162602349] A211212
         A030662,A211213,  A181991,
For example the number of ordered set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] are 1, 168, 1680 respectively. Thus A(3,3) = -1 + 168 - 1680 = -1513.
Formatted as a triangle:
[1]
[1, -1]
[1, -1,  1]
[1, -1,  1,    -5]
[1, -1,  5,    -1,   21]
[1, -1, 19,   -61,    1, -105]
[1, -1, 69, -1513, 1385,   -1, 635]

Cf. A002115, A028296, A030662, A033999, A181991, A211212, A211213, A260845, A260833, A260875, A260876.

def A260877(m,n):
    shapes = ([x*m for x in p] for p in Partitions(n).list())
    return sum((-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s). cardinality() for s in shapes)
for m in (0..5): print([A260877(m,n) for n in (0..7)])

A318148 Coefficients of the Omega polynomials of order 4, triangle T(n,k) read by rows with 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, -34, 35, 0, 11056, -16830, 5775, 0, -14873104, 27560780, -15315300, 2627625, 0, 56814228736, -119412815760, 84786627900, -24734209500, 2546168625, 0, -495812444583424, 1140896479608800, -948030209181000, 364143337057500, -65706427536750, 4509264634875

Offset: 0

Peter Luschny, Aug 22 2018

The name 'Omega polynomial' is not a standard name.

			[0] [1]
[1] [0,           1]
[2] [0,         -34,            35]
[3] [0,       11056,        -16830,        5775]
[4] [0,   -14873104,      27560780,   -15315300,      2627625]
[5] [0, 56814228736, -119412815760, 84786627900, -24734209500, 2546168625]

All row sums are 1, alternating row sums (taken absolute) are A211212.
T(n,1) ~ A273352(n), T(n,n) = A025036(n).
A023531 (m=1), A318146 (m=2), A318147 (m=3), this seq (m=4).

# See A318146 for the missing functions.
FL([seq(CL(OmegaPolynomial(4, n)), n=0..8)]);

(* OmegaPolynomials are defined in A318146 *)
Table[CoefficientList[OmegaPolynomial[4, n], x], {n, 0, 6}] // Flatten

# See A318146 for the function OmegaPolynomial.
[list(OmegaPolynomial(4, n)) for n in (0..6)]

Omega(m, n, z) = (m*n)!*[z^(n*m)] H(m, z)^x where H(m, z) = hypergeom([], [seq(i/m, i=1..m-1)], (z/m)^m). We consider here the case m = 4 (for other cases see the cross-references).

cp's OEIS Frontend

A278074 Triangle read by rows, coefficients of the polynomials P(m, n) = Sum_{k=1..n} binomial(m*n, m*k)* P(m, n-k)*z with P(m, 0) = 1 and m = 4.

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A181985 Generalized Euler numbers. Square array A(n,k), n >= 1, k >= 0, read by antidiagonals. A(n,k) = n-alternating permutations of length n*k.

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A292604 Triangle read by rows, coefficients of generalized Eulerian polynomials F_{2}(x).

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A292606 Triangle read by rows, coefficients of generalized Eulerian polynomials F_{4;n}(x).

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A327024 Ordered set partitions of the set {1, 2, ..., 4*n} with all block sizes divisible by 4, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.

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A260877 Square array read by ascending antidiagonals: number of m-shape Euler numbers.

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A318148 Coefficients of the Omega polynomials of order 4, triangle T(n,k) read by rows with 0<=k<=n.

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A278074 Triangle read by rows, coefficients of the polynomials P(m, n) = Sum_{k=1..n} binomial(mn, mk)* P(m, n-k)*z with P(m, 0) = 1 and m = 4.