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

1,1

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

Prog

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

Crossrefs

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

Sequence in context: A368332 A224765 A369167 * A115621 A326514 A077565

Adjacent sequences: A160264 A160265 A160266 * A160268 A160269 A160270

Keyword

nonn

Author

Vladimir Shevelev, May 07 2009, May 11 2009

Extensions

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

Status

approved