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a(n) = lcm(1..Catalan(n)).
4

%I #19 Nov 21 2023 18:13:22

%S 1,1,2,60,360360,219060189739591200,

%T 1749342047920660916901891145781670987072592322134428432000

%N a(n) = lcm(1..Catalan(n)).

%C For every cycle count LCM-sequence Axxxxxx in A073204 it holds that Axxxxxx(n) divides a(n). And this also applies to similar LCM-sequences induced by other "Catalan bijections", cf. A060113.

%C The next term (a(7)) has 184 digits. - _Harvey P. Dale_, Nov 21 2023

%H Muniru A Asiru, <a href="/A078491/b078491.txt">Table of n, a(n) for n = 0..7</a>

%H <a href="/index/Lc#lcm">Index entries for sequences related to lcm's</a>

%F a(n) = A003418(A000108(n)).

%t Table[LCM@@Range[CatalanNumber[n]],{n,0,7}] (* _Harvey P. Dale_, Nov 21 2023 *)

%o (PARI) a(n) = lcm([1..binomial(2*n,n)/(n+1)]); \\ _Michel Marcus_, Mar 21 2018

%o (GAP) List([0..7],n->Lcm([1..Binomial(2*n,n)/(n+1)])); # _Muniru A Asiru_, Mar 21 2018

%Y Composition of the sequences A000108 and A003418.

%K nonn

%O 0,3

%A _Antti Karttunen_, Jan 04 2003