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A160267 Minimum of A122458(n) and A160266(n).

%I A160267 #13 Sep 23 2018 20:58:26
%S A160267 2,1,1,1,3,1,1,1,2,1,2,1,1,1,1,1,1,1,1,1,3,1,2,1,1,1,1,1,4,1,1,1,2,1,
%T A160267 1,1,1,1,2,1,2,1,1,1,1,1,5,1,1,1,17,1,3,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,
%U A160267 3,1,4,1,2,1,1,1,1,1,1,1,2,1,9,1,2,1,1,1,2,1,3,1,1,1,8,1,1,1,5,1,2,1,1,1,2,1
%N A160267 Minimum of A122458(n) and A160266(n).
%C A160267 Let f be the operation defined in A159885, namely f(2n+1) = A075677(n+1), and f^k its k-fold iteration.
%C A160267 Then a(n) is the smallest k such that either f^k(2n+1)< 2n+1 or A006694((f^k(2n+1)-1)/2) < A006694(n).
%H A160267 Antti Karttunen, <a href="/A160267/b160267.txt">Table of n, a(n) for n = 1..65537</a>
%H A160267 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%o A160267 (PARI)
%o A160267 f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined for odd n only. Cf. A075677.
%o A160267 A006519(n) = (1<<valuation(n, 2));
%o A160267 A006694(n) = (sumdiv(2*n+1, d, eulerphi(d)/znorder(Mod(2, d))) - 1); \\ From A006694
%o A160267 A160267(n) = { my(w=A006694(n), u = (n+n+1), k=0); while((u >= (n+n+1))&&(A006694((u-1)/2) >= w), k++; u = f(u)); (k); }; \\ _Antti Karttunen_, Sep 22 2018
%Y A160267 Cf. A006694, A075677, A122458, A159885, A159945, A160198, A160266.
%K A160267 nonn
%O A160267 1,1
%A A160267 _Vladimir Shevelev_, May 07 2009, May 11 2009
%E A160267 a(47) corrected and more terms appended by _R. J. Mathar_, Aug 08 2010