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%I #48 Oct 27 2023 21:55:01
%S 1,1,1,3,2,10,15,35,28,252,210,2310,1980,1716,3003,45045,40040,680680,
%T 612612,554268,503880,10581480,9699690,44618574,41186376,114406600,
%U 106234700,2868336900,2677114440,77636318760,145568097675,136745788725
%N a(n) = lcm{1, 2, ..., n}/(n*(n-1)), n >= 2.
%H Alois P. Heinz, <a href="/A099946/b099946.txt">Table of n, a(n) for n = 2..1000</a> (first 99 terms from Luke March)
%F a(n) = A003418(n)/(n*(n-1)) = A003418(n)/A002378(n-1), n >= 2.
%p a:= n-> ilcm(seq(k,k=1..n))/n/(n-1): seq(a(n), n=2..37); # _Emeric Deutsch_, Jun 13 2005
%t Table[LCM@@Range[n]/(n(n-1)), {n,2,40}] (* _Harvey P. Dale_, Jan 14 2011 *)
%o (Python)
%o from math import gcd
%o def lcm(a, b):
%o return (a * b) // gcd(a, b)
%o def f(lim):
%o l = 1
%o for n in range(2, lim + 1):
%o l = lcm(n, l)
%o print(n, l // (n * (n - 1)))
%o f(100) # _Luke March_, Jul 23 2014
%o (PARI) a(n) = lcm(vector(n, i, i))/(n*(n-1)); \\ _Michel Marcus_, Jul 25 2014
%Y Cf. A002378, A002944, A003418, A025558.
%K nonn,easy
%O 2,4
%A _N. J. A. Sloane_, Nov 12 2004
%E More terms from _Emeric Deutsch_, Jun 13 2005
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