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Let f(x)=|log x|, f^k(x)=f(f^(k-1)(x))=f iterated k times at x; a(n) = least k such that f^k(n) <= 1/n.
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%I #9 Jul 21 2015 00:29:12

%S 1,2,2,4,6,9,5,27,7,30,8,25,3,3,3,3,3,6,9,27,52,43,8,19,17,97,56,24,

%T 193,131,24,27,128,87,7,10,17,118,111,116,38,26,9,104,36,61,24,13,60,

%U 17,37,10,21,140,48,48,13,19,243,8,33,25,171,25,18,20,20,5,5,5,5,90,55,47,44

%N Let f(x)=|log x|, f^k(x)=f(f^(k-1)(x))=f iterated k times at x; a(n) = least k such that f^k(n) <= 1/n.

%e |log 4| ~= 1.38629 > 1/4

%e |log |log 4| | ~= 0.32663 > 1/4

%e |log |log |log 4| | | ~= 1.11891 > 1/4

%e |log |log |log |log 4| | | | ~= 0.11235 < 1/4. Thus a(4)=4

%t f[n_] := N[ Abs[ Log[ n]], 1024]; g[n_] := Block[{k = 1}, While[ N[ Nest[f, n, k], 64] > 1/n, k++ ]; k]; Table[ g[n], {n, 75}] (* _Robert G. Wilson v_, Mar 16 2004 *)

%Y First occurrence in A093079, records in A093081 and their index in A093080.

%K nonn

%O 1,2

%A J. Taylor (integersfan(AT)yahoo.com), Mar 14 2004

%E More terms from _Robert G. Wilson v_, Mar 16 2004