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A034888 Number of digits in 3^n.

%I A034888 #29 Dec 23 2022 17:25:50
%S A034888 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,11,12,12,13,13,14,
%T A034888 14,15,15,16,16,17,17,18,18,19,19,20,20,21,21,21,22,22,23,23,24,24,25,
%U A034888 25,26,26,27,27,28,28,29,29,30,30,31,31,32,32,32,33,33
%N A034888 Number of digits in 3^n.
%H A034888 G. C. Greubel, <a href="/A034888/b034888.txt">Table of n, a(n) for n = 0..10000</a>
%F A034888 a(n) = floor(n*log(3)/log(10)) + 1. E.g., a(10)=5 because 3^10 = 59049 and floor(10*log(3)/log(10)) + 1 = 4 + 1 = 5. - _Jaap Spies_, Dec 15 2003
%F A034888 a(n) = A055642(3^n) = A055642(A000244(n)). - _Michel Marcus_, Jun 23 2015
%p A034888 seq(floor(n*ln(3)/ln(10))+1, n=0..100);
%t A034888 Table[Length[IntegerDigits[3^n]], {n, 0, 100}] (* _T. D. Noe_, Mar 20 2012 *)
%t A034888 IntegerLength[3^Range[0,70]] (* _Harvey P. Dale_, Jan 28 2015 *)
%o A034888 (PARI) a(n)=#Str(3^n) \\ _Charles R Greathouse IV_, Jul 03 2013
%o A034888 (Magma) [#Intseq(3^n): n in [0..100] ]; // _Vincenzo Librandi_, Jun 23 2015
%o A034888 (Python)
%o A034888 def a(n): return len(str(3**n))
%o A034888 print([a(n) for n in range(70)]) # _Michael S. Branicky_, Dec 23 2022
%Y A034888 Cf. A000244 (powers of 3), A055642 (number of digits of n).
%K A034888 nonn,base,easy
%O A034888 0,4
%A A034888 _N. J. A. Sloane_