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Number of digits in (2n)! / (2^n * n!).
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%I #31 Jan 15 2024 14:57:16

%S 1,1,2,3,3,5,6,7,8,9,11,12,13,15,16,18,19,21,22,24,26,27,29,31,32,34,

%T 36,37,39,41,43,45,46,48,50,52,54,56,58,59,61,63,65,67,69,71,73,75,77,

%U 79,81,83,85,87,89,91,93,95,97,99,101,104,106,108,110

%N Number of digits in (2n)! / (2^n * n!).

%C a(n) is the number of digits in double factorial of odd numbers (see A001147).

%H Robert Israel, <a href="/A338375/b338375.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = floor(log((2n)!/((2^n)*n!))/log(10))+1.

%F a(n) = A055642(A001147(n)).

%e For n = 7, (2*n-1)!! = 13!! = 135135 and the number of digits is 6.

%p seq(1 + ilog10(doublefactorial(2*n-1)), n=1..100); # _Robert Israel_, Jan 15 2024

%t a[n_] := IntegerLength[(2 n - 1)!!]; Array[a, 65] (* _Amiram Eldar_, Oct 23 2020 *)

%o (Sage) floor((log(factorial(2 * n) / ((2 ** n) * factorial(n))) / log(10))) + 1

%Y Cf. A001147, A055642, A068398.

%K nonn,base

%O 1,3

%A _Jeremias M. Gomes_, Oct 23 2020