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a(n) = lcm(1, 2, ..., 2n) / lcm(1, 2, ..., n).
4

%I #22 May 30 2019 11:19:40

%S 2,6,10,70,42,462,858,858,4862,92378,8398,193154,74290,222870,6463230,

%T 200360130,11785890,11785890,22951470,22951470,941010270,40463441610,

%U 1759280070,82686163290,115760628606,115760628606,2045104438706

%N a(n) = lcm(1, 2, ..., 2n) / lcm(1, 2, ..., n).

%C Also, lcm(n+1, n+2, ..., 2n-1, 2n) / lcm(1, 2, ..., n-1, n).

%H T. D. Noe, <a href="/A093880/b093880.txt">Table of n, a(n) for n = 1..500</a>

%H J. Sondow, <a href="https://doi.org/10.1090/S0002-9939-03-07081-3">Criteria for irrationality of Euler's constant</a>, Criteria for irrationality of Euler's constant, Proc. Amer. Math. Soc. 131 (2003) 3335-3344.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LeastCommonMultiple.html">Least Common Multiple</a>

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

%F The prime number theorem implies that a(n) = e^(n(1+o(1))) as n -> infinity. In other words, log(a(n))/n -> 1 as n -> infinity. - _Jonathan Sondow_, Jan 17 2005

%F a(n) = A003418(2n)/A003418(n) = A099996(n)/A003418(n).

%e The LCM of {1,2,3,4,5,6} is 60 and the LCM of {1,2,3} is 6, so a(3) = 60/6 = 10.

%p a:=n->lcm(seq(j,j=n+1..2*n))/lcm(seq(j,j=1..n)): seq(a(n),n=1..32); # _Emeric Deutsch_, Feb 02 2006

%t f[n_] := LCM @@ Table[i, {i, 2n}]/LCM @@ Table[i, {i, n}]; Table[ f[n], {n, 27}] (* _Robert G. Wilson v_, Jan 22 2005 *)

%Y Cf. A080397.

%K nonn

%O 1,1

%A _Amarnath Murthy_, Apr 22 2004

%E More terms from _Emeric Deutsch_, Feb 02 2006

%E Entry revised by _N. J. A. Sloane_, Jan 24 2007