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Number of digits in (2^n)*(n!).
1

%I #14 May 03 2024 22:58:57

%S 1,1,1,2,3,4,5,6,8,9,10,11,13,14,16,17,19,20,22,23,25,27,28,30,32,33,

%T 35,37,38,40,42,44,46,47,49,51,53,55,57,59,60,62,64,66,68,70,72,74,76,

%U 78,80,82,84,86,88,90,92,94,96,98,100,103,105,107,109,111,113,115,117,120

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

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

%F a(n) = A055642(A000165(n)). - _Michel Marcus_, Oct 23 2020

%t Table[IntegerLength[2^n n!],{n,0,70}] (* _Harvey P. Dale_, May 03 2024 *)

%o (PARI) a(n) = log(n!*2^n)\log(10) + 1; \\ _Michel Marcus_, Oct 23 2020

%Y Cf. A000165, A055642.

%K nonn,base

%O 0,4

%A _Benoit Cloitre_, Mar 01 2002

%E a(0)=1 prepended by _Alois P. Heinz_, Oct 23 2020