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%I #24 Oct 22 2021 12:03:45
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,27,
%T 28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,
%U 52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75
%N Number of digits in 11^n.
%H Seiichi Manyama, <a href="/A348553/b348553.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A055642(A001020(n)) = A055642(11^n).
%e a(24) = 25 because 11^24 = 9849732675807611094711841, which has 25 digits.
%e a(25) = 27 because 11^25 = 108347059433883722041830251, which has 27 digits.
%t a[n_] := IntegerLength[11^n]; Array[a, 100, 0] (* _Amiram Eldar_, Oct 22 2021 *)
%o (PARI) a(n) = #Str(11^n);
%o (Python)
%o def a(n): return len(str(11**n))
%o print([a(n) for n in range(98)]) # _Michael S. Branicky_, Oct 22 2021
%Y Number of digits in b^n: A034887 (b=2), A034888 (b=3), A210434 (b=4), A210435 (b=5), A210436 (b=6), A210062 (b=7), this sequence (b=11).
%Y Cf. A001020, A055642.
%K nonn,easy,base
%O 0,2
%A _Seiichi Manyama_, Oct 22 2021
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