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

%I A359365 #14 Dec 30 2022 11:23:53
%S A359365 1,1,2,6,72,240,3600,75600,1411200,10160640,457228800,4191264000,
%T A359365 184415616000,2054916864000,12466495641600,1308982042368000,
%U A359365 314155690168320000,14241724620963840000,2178983867007467520000,37260624125827694592000,337119932567012474880000
%N A359365 a(n) = lcm([ n!*binomial(n-1, m-1) / m! for m = 1..n ]) with a(0) = 1.
%C A359365 The lcm of the rows of the unsigned Lah triangle (for k >= 1).
%H A359365 Michael De Vlieger, <a href="/A359365/b359365.txt">Table of n, a(n) for n = 0..390</a>
%p A359365 # Maple has the convention integer lcm() = 1, which covers the case n = 0.
%p A359365 a := n -> ilcm(seq(n!*binomial(n-1, m-1) / m!, m = 1..n)):
%p A359365 seq(a(n), n = 0..20);
%t A359365 {1}~Join~Table[LCM @@ Array[n!*Binomial[n - 1, # - 1]/#! &, n], {n, 20}] (* _Michael De Vlieger_, Dec 30 2022 *)
%o A359365 (Python)
%o A359365 from functools import cache
%o A359365 from sympy import lcm
%o A359365 def A359365 (n: int) -> int:
%o A359365     @cache
%o A359365     def l(n: int) -> list[int]:
%o A359365         if n == 0: return [1]
%o A359365         row: list[int] = l(n - 1) + [1]
%o A359365         row[0] = 0
%o A359365         for k in range(n - 1, 0, -1):
%o A359365             row[k] = row[k] * (n + k - 1) + row[k - 1]
%o A359365         return row
%o A359365     return lcm(l(n)[1:])
%o A359365 print([A359365(n) for n in range(21)])
%o A359365 (PARI) a(n) = lcm(vector(n, m, n!*binomial(n-1, m-1) / m!)); \\ _Michel Marcus_, Dec 30 2022
%Y A359365 Cf. A271703 (unsigned Lah numbers), A103505 (gcd counterpart).
%K A359365 nonn
%O A359365 0,3
%A A359365 _Peter Luschny_, Dec 30 2022