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%I #14 Oct 27 2023 22:05:37
%S 1,4,5,5,6,7,8,8,9,9,10,11,11,12,12,13,14,14,15,15,16,16,17,17,18,18,
%T 19,19,20,20,21,21,22,22,23,23,23,24,24,25,25,26,26,27,27,28,28,28,29,
%U 29,30,30,31,31,32,32,32,33,33,34,34,35,35,35,36,36,37,37,37
%N Smallest positive m such that n^m has at least n digits.
%F a(n) = ceiling((n-1)/log_10(n)), n > 1. - _Vladeta Jovovic_, Dec 23 2001
%e a(15) = 12, as 15^12 = 129746337890625 has 15 digits, while 15^11 = 8649755859375 has only 13 digits.
%p f := []: for i from 1 to 50 do for j from 1 do if length(i^j) >= i then f := [op(f),j]; break fi; od; od; op(f);
%t Join[{1},Table[Ceiling[(n-1)/Log[10,n]],{n,2,70}]] (* _Harvey P. Dale_, Apr 14 2016 *)
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, May 06 2001
%E Corrected and extended by _Asher Auel_, May 14 2001
%E Offset corrected by _Sean A. Irvine_, Feb 18 2023
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