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The number of digits of (2^n)!.
2

%I #15 Mar 25 2024 00:15:29

%S 1,1,2,5,14,36,90,216,507,1167,2640,5895,13020,28504,61937,133734,

%T 287194,613842,1306594,2771010,5857670,12346641,25955890,54436999,

%U 113924438,237949763,496101303,1032606162,2146019444,4453653132,9230534755

%N The number of digits of (2^n)!.

%t LogBase10Stirling[n_] := Floor[ Log[10, 2 Pi n]/2 + n*Log[10, n/E] + Log[10, 1 + 1/(12 n) + 1/(288 n^2) - 139/(51840 n^3) - 571/(2488320 n^4) + 163879/(209018880 n^5)]]; Table[ LogBase10Stirling[2^n] + 1, {n, 0, 30}]

%t IntegerLength[(2^Range[0,30])!] (* _Harvey P. Dale_, Nov 05 2021 *)

%Y Cf. A000722, A061010, A244060.

%K nonn,base

%O 0,3

%A _Robert G. Wilson v_, Jun 18 2014