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A160364 Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1)) < A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2) < A006694(n)} 0
2, 1, 1, 5, 3, 1, 1, 2, 5, 1, 2, 1, 1, 1, 1, 5, 2, 5, 3, 33, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 5, 10, 1, 1, 2, 5, 5, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n.

LINKS

Table of n, a(n) for n=1..43.

EXAMPLE

Beginning with n=1, we have f(2n+1)=f(3)=5. Here A000120(3)=A000120(5)=2 and A006694((3-1)/2)= A006694((5-1)/2)=1. None of values did not become less than. Therefore a(1)>1. Since f(5)=1 and A000120(1)=1 and A006694(0)=0, then a(2)=2.

CROSSREFS

A000120 A006694 A160267 A006694 A122458 A160266 A159885 A159945 A160198

Sequence in context: A011281 A300731 A100398 * A107735 A137570 A079213

Adjacent sequences:  A160361 A160362 A160363 * A160365 A160366 A160367

KEYWORD

nonn,uned

AUTHOR

Vladimir Shevelev, May 11 2009

STATUS

approved

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Last modified December 6 04:20 EST 2021. Contains 349562 sequences. (Running on oeis4.)