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

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, 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, 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

Offset: 1

Author

Vladimir Shevelev, May 07 2009, May 11 2009

Keywords

• nonn

Comments

Let f be the operation defined in A159885, namely f(2n+1) = A075677(n+1), and f^k its k-fold iteration.

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).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006694, A075677, A122458, A159885, A159945, A160198, A160266.

Programs

PARI

f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined for odd n only. Cf. A075677.
A006519(n) = (1<A006694(n) = (sumdiv(2*n+1, d, eulerphi(d)/znorder(Mod(2, d))) - 1); \\ From A006694
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

Extensions

a(47) corrected and more terms appended by R. J. Mathar, Aug 08 2010