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A160267 Minimum of A122458(n) and A160266(n). 7
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 (list; graph; refs; listen; history; text; internal format)
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: A079229 A204988 A224765 * 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

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Last modified December 5 17:42 EST 2019. Contains 329768 sequences. (Running on oeis4.)