A271703 -id:A271703 - cp's OEIS frontend

A216154 Triangle read by rows, T(n,k) n>=0, k>=0, generalization of A000255.

1, 1, 1, 3, 4, 1, 11, 21, 9, 1, 53, 128, 78, 16, 1, 309, 905, 710, 210, 25, 1, 2119, 7284, 6975, 2680, 465, 36, 1, 16687, 65821, 74319, 35035, 7945, 903, 49, 1, 148329, 660064, 857836, 478464, 133630, 19936, 1596, 64, 1, 1468457, 7275537, 10690812, 6879684, 2279214, 419958, 44268, 2628, 81, 1

Offset: 0

Peter Luschny, Sep 19 2012

			     1,
     1,      1,
     3,      4,      1,
    11,     21,      9,      1,
    53,    128,     78,     16,      1,
   309,    905,    710,    210,     25,      1,
  2119,   7284,   6975,   2680,    465,     36,      1,
16687,  65821,  74319,  35035,   7945,    903,     49,      1,
148329, 660064, 857836, 478464, 133630,  19936,   1596,     64,      1,

A000255 (col. 0), A110450 (diag. n,n-2).

Cf. A111596, A271703.

A216154 := proc(n,k) local L, Z;
L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*C(n,n-k)*C(n-1,n-k)):
Z := (n,k) -> add(C(-j,-n)*L(j,k), j=0..n);
Z(n+1, k+1) end:
seq(seq(A216154(n,k), k=0..n), n=0..9); # Peter Luschny, Apr 13 2016

T[0, 0] = 1; T[0, ] = 0; T[n, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (2k+1) T[n-1, k] + (k+1) (k+2) T[n-1, k+1]; T[, ] = 0;
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 02 2019 *)

def A216154_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+(1+2*k)*M[n-1,k]+(k+1)*(k+2)*M[n-1,k+1]
    return M
A216154_triangle(9)

Recurrence: T(0,0)=1, T(0,k)=0 for k>0 and for n>=1 T(n,k) = T(n-1,k-1)+(1+2*k)*T(n-1,k)+(k+1)*(k+2)*T(n-1,k+1).

Let Z(n, k) = Sum_{j=0..n} C(-j, -n)*L(j, k) where L denotes the unsigned Lah numbers A271703. Then T(n, k) = Z(n+1, k+1). - Peter Luschny, Apr 13 2016

A330609 T(n, k) = binomial(n-k-1, k-1)*(n-k)!/k! for n >= 0 and 0 <= k <= floor(n/2). Irregular triangle read by rows.

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 0, 6, 1, 0, 24, 6, 0, 120, 36, 1, 0, 720, 240, 12, 0, 5040, 1800, 120, 1, 0, 40320, 15120, 1200, 20, 0, 362880, 141120, 12600, 300, 1, 0, 3628800, 1451520, 141120, 4200, 30, 0, 39916800, 16329600, 1693440, 58800, 630, 1

Offset: 0

Peter Luschny, Dec 27 2019

Also the antidiagonals of the Lah triangle A271703.

			Triangle begins:
[0] 1
[1] 0
[2] 0, 1
[3] 0, 2
[4] 0, 6,     1
[5] 0, 24,    6
[6] 0, 120,   36,    1
[7] 0, 720,   240,   12
[8] 0, 5040,  1800,  120,  1
[9] 0, 40320, 15120, 1200, 20

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Variants: A180047, A221913. Row sums: A001053.

Cf. A271703.

T := (n, k) -> binomial(n-k-1, k-1)*(n-k)!/k!:
seq(seq(T(n, k), k=0..floor(n/2)), n=0..12);
# Alternative:
T := proc(n, k) option remember;
if (n=0 and k=0) or (n=2 and k=1) then 1 elif (k < 1) or (k > ceil(n/2)) then 0
else (n-1)*T(n-1, k) + T(n-2, k-1) fi end: seq(seq(T(n, k), k=0..n/2), n=0..12);

Table[Binomial[n-k-1,k-1] (n-k)!/k!,{n,0,20},{k,0,Floor[n/2]}]//Flatten (* Harvey P. Dale, Oct 19 2021 *)

T(0,0) = T(2,1) = 1. If k < 1 or k > ceiling(n/2) then T(n,k) = 0. Otherwise:

T(n, k) = (n-1)*T(n-1, k) + T(n-2, k-1)

A355005 Table read by rows. T(n, k) = n((k + n)!)^2/((k + n)(n!)^2*k!) for n > 0 and T(0, 0) = 1.

Original entry on oeis.org

1, 1, 2, 1, 6, 36, 1, 12, 120, 1200, 1, 20, 300, 4200, 58800, 1, 30, 630, 11760, 211680, 3810240, 1, 42, 1176, 28224, 635040, 13970880, 307359360, 1, 56, 2016, 60480, 1663200, 43908480, 1141620480, 29682132480, 1, 72, 3240, 118800, 3920400, 122316480, 3710266560, 111307996800, 3339239904000

Offset: 0

Peter Luschny, Jun 15 2022

			[0] 1;
[1] 1,  2;
[2] 1,  6,   36;
[3] 1, 12,  120,  1200;
[4] 1, 20,  300,  4200,   58800;
[5] 1, 30,  630, 11760,  211680,  3810240;
[6] 1, 42, 1176, 28224,  635040, 13970880,  307359360;
[7] 1, 56, 2016, 60480, 1663200, 43908480, 1141620480, 29682132480;

T(n, 1) = A002378, T(n, n) = A187535, A355004 (row sums), A271703 (Lah).

T := (n, k) -> ifelse(n = 0, 1, n*((k + n)!)^2 / ((k + n)*(n!)^2*k!)):
seq(seq(T(n, k), k = 0..n), n = 0..8);

T(n, k) = Lah(k + n, n), where Lah denotes the unsigned Lah numbers A271703.

A359365 a(n) = lcm([ n!*binomial(n-1, m-1) / m! for m = 1..n ]) with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 6, 72, 240, 3600, 75600, 1411200, 10160640, 457228800, 4191264000, 184415616000, 2054916864000, 12466495641600, 1308982042368000, 314155690168320000, 14241724620963840000, 2178983867007467520000, 37260624125827694592000, 337119932567012474880000

Offset: 0

Peter Luschny, Dec 30 2022

nonn

The lcm of the rows of the unsigned Lah triangle (for k >= 1).

Michael De Vlieger, Table of n, a(n) for n = 0..390

Cf. A271703 (unsigned Lah numbers), A103505 (gcd counterpart).

# Maple has the convention integer lcm() = 1, which covers the case n = 0.
a := n -> ilcm(seq(n!*binomial(n-1, m-1) / m!, m = 1..n)):
seq(a(n), n = 0..20);

{1}~Join~Table[LCM @@ Array[n!*Binomial[n - 1, # - 1]/#! &, n], {n, 20}] (* Michael De Vlieger, Dec 30 2022 *)

a(n) = lcm(vector(n, m, n!*binomial(n-1, m-1) / m!)); \\ Michel Marcus, Dec 30 2022

from functools import cache
from sympy import lcm
def A359365 (n: int) -> int:
    @cache
    def l(n: int) -> list[int]:
        if n == 0: return [1]
        row: list[int] = l(n - 1) + [1]
        row[0] = 0
        for k in range(n - 1, 0, -1):
            row[k] = row[k] * (n + k - 1) + row[k - 1]
        return row
    return lcm(l(n)[1:])
print([A359365(n) for n in range(21)])

A343581 a(n) = binomial(n, floor(n/2))*FallingFactorial(n - 1, n - floor(n/2)).

Original entry on oeis.org

1, 0, 2, 6, 36, 240, 1200, 12600, 58800, 846720, 3810240, 69854400, 307359360, 6849722880, 29682132480, 779155977600, 3339239904000, 100919250432000, 428906814336000, 14668613050291200, 61934143990118400, 2364758225077248000, 9931984545324441600, 418798681661180620800

Offset: 0

Peter Luschny, Apr 21 2021

nonn

Partially ordered sets on n elements that consist entirely of floor(n/2) chains (nonempty, linearly ordered subsets).

Cf. A187535, A271703.

a := n -> `if`(n=0, 1, binomial(n - 1, iquo(n,2) - 1)*n!/iquo(n, 2)!):
seq(a(n), n = 0..21);

a(n) = sum(j=n\2, n, abs(stirling(n, j, 1))*stirling(j, n\2, 2)); \\ Michel Marcus, Apr 22 2021

def a(n): return binomial(n, n - n//2)*falling_factorial(n - 1, n - n//2)
print([a(n) for n in range(22)])

a(n) = Sum_{j=floor(n/2)..n} |Stirling1(n, j)|*Stirling2(j, floor(n/2)).
a(n) = binomial(n - 1, floor(n/2) - 1)*n!/floor(n/2)! for n >= 1, a(0) = 1.
a(n) = A271703(n, floor(n/2)).

A367776 a(n) = binomial(2n, n - 1)(2*n + 1)! / n!.

Original entry on oeis.org

0, 6, 240, 12600, 846720, 69854400, 6849722880, 779155977600, 100919250432000, 14668613050291200, 2364758225077248000, 418798681661180620800, 80831074222717378560000, 16887920864389166592000000, 3797443866983262444748800000, 914438045469094536918528000000

Offset: 0

Peter Luschny, Nov 29 2023

nonn

Cf. A000108 (Catalan), A271703.

seq(binomial(2*n, n - 1)*(2*n + 1)! / n!, n = 0..15);

a[n_]:=n*CatalanNumber[n]*Gamma[2*n+2]/n!;Flatten[Table[a[n],{n,0,15}]] (* Detlef Meya, Dec 02 2023 *)

a(n) = A271703(2*n + 1, n).
a(n) = binomial(2*n+1,n)*(2n)!/(n-1)! for n > 0. - Chai Wah Wu, Nov 30 2023
a(n) = n*A000108(n)*(2*n + 1)!/n!. - Detlef Meya, Dec 02 2023

cp's OEIS Frontend

A216154 Triangle read by rows, T(n,k) n>=0, k>=0, generalization of A000255.

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Formula

A330609 T(n, k) = binomial(n-k-1, k-1)*(n-k)!/k! for n >= 0 and 0 <= k <= floor(n/2). Irregular triangle read by rows.

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Maple

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Formula

A355005 Table read by rows. T(n, k) = n*((k + n)!)^2/((k + n)*(n!)^2*k!) for n > 0 and T(0, 0) = 1.

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Maple

Formula

A359365 a(n) = lcm([ n!*binomial(n-1, m-1) / m! for m = 1..n ]) with a(0) = 1.

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PARI

Python

A343581 a(n) = binomial(n, floor(n/2))*FallingFactorial(n - 1, n - floor(n/2)).

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Programs

Maple

PARI

SageMath

Formula

A367776 a(n) = binomial(2*n, n - 1)*(2*n + 1)! / n!.

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Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Formula

A355005 Table read by rows. T(n, k) = n((k + n)!)^2/((k + n)(n!)^2*k!) for n > 0 and T(0, 0) = 1.

A367776 a(n) = binomial(2n, n - 1)(2*n + 1)! / n!.