%I #24 Mar 07 2022 18:17:49
%S 1,4,9,48,50,360,735,2240,2268,25200,25410,332640,334620,336336,
%T 675675,11531520,11571560,220540320,221152932,221707200,222211080,
%U 5121436320,5131136010,25700298624,25741485000,77338861600,77445096300,2248776129600,2251453244040
%N a(n) = (n/(n+1)) * lcm(1,2,...,n+1).
%C a(n) = (1/1 + 1/3 + 1/6 + ... + 1/C(n+1,2))*lcm(1,3,6,...,binomial(n+1,2)) = 2n/(n+1) * lcm(1,3,6,...,binomial(n+1,2)).
%C a(n+1) = a(n) * ((n+1)^2)/(n * ((n+2)/p) ), where p = n+2 if n+2 is prime, p = q if n+2 = q^k (q is prime, k>1), or p = 1 if n+2 is not a prime or a prime power. - Scott C. Macfarlan (scottmacfarlan(AT)covance.com), Jan 08 2004
%F a(n) = n * A002944(n+1) = (n/(n+1)) * A003418(n+1).
%p a:= n-> (n/(n+1)) * ilcm($1..n+1):
%p seq(a(n), n=1..29); # _Alois P. Heinz_, Mar 07 2022
%t Table[n/(n+1) LCM@@Range[n+1],{n,30}] (* _Harvey P. Dale_, Apr 02 2011 *)
%o (PARI) a(n) = n*lcm([1..n+1])/(n+1); \\ _Michel Marcus_, Mar 07 2022
%Y Cf. A002944, A003418.
%K nonn
%O 1,2
%A _Clark Kimberling_
%E Entry revised by _N. J. A. Sloane_, Nov 12 2004