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

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

Offset: 0

Author

Peter Luschny, Dec 30 2022

Keywords

• nonn

Comments

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

Links

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

Crossrefs

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

Programs

Maple

# 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);

Mathematica

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

PARI

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

Python

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)])