A159885 - OEIS

A159885

For n >= 1, let f(2n+1) = (3n+2)/A006519(3n+2) and let f^k be the k-th iteration of f. Then a(n) is the least k such that A000120(f^k(2n+1)) <= A000120(n).

2, 1, 2, 6, 1, 1, 2, 3, 3, 1, 1, 4, 1, 1, 2, 8, 2, 3, 3, 39, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 2, 8, 5, 2, 2, 41, 3, 2, 3, 5, 5, 1, 1, 1, 1, 1, 1, 42, 2, 1, 4, 6, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 44, 5, 5, 5, 31, 5, 2, 2, 41, 7, 1, 3, 3, 3, 2, 3, 34, 3, 5, 13, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 42, 8, 1, 2, 4, 1

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OFFSET

1,1

COMMENTS

Conjecture: a(n) exists for every n >= 1. It is easy to see that this conjecture is equivalent to the well-known Collatz 3x+1 conjecture.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to 3x+1 (or Collatz) problem

PROG

(PARI)

A006519(n) = (1<<valuation(n, 2));

f(n) = ((3*((n-1)/2))+2)/A006519((3*((n-1)/2))+2); \\ Defined only for odd n. Cf. A075677.