

A135730


Number of steps to reach the minimum of the final cycle under iterations of the map A001281: x>3x1 if x odd, x/2 otherwise.


8



0, 1, 4, 2, 0, 5, 3, 3, 11, 1, 6, 6, 9, 4, 9, 4, 0, 12, 7, 2, 8, 7, 3, 7, 16, 10, 5, 5, 10, 10, 6, 5, 19, 1, 13, 13, 14, 8, 13, 3, 9, 9, 8, 8, 22, 4, 16, 8, 17, 17, 11, 11, 16, 6, 12, 6, 29, 11, 11, 11, 7, 7, 19, 6, 37, 20, 20, 2, 19, 14, 19, 14, 15, 15, 9, 9, 14, 14, 14, 4
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OFFSET

1,3


COMMENTS

Under iterations of the map A001281, the orbit of any positive integer seems to end in one of 3 possible cycles, having 1, 5, resp. 17 as smallest element. This sequence gives the number of iterations needed to reach one of these values. Another sequence that could be considered is the number of iterations needed to reach /any/ element of the final cycle.
From N. J. A. Sloane, Sep 04 2015: (Start)
The same sequence arises as follows: Start at 2n1 and repeatedly apply the map (see A261671): subtract 1 and divide by 2 if the result is odd, otherwise multiply by 3; a(n) is the number of steps to reach one of 1, 9, or 33.
It is conjectured that the trajectory of any odd number will eventually reach 1, 9, or 33, and so enter one of the loops (1,3), (9, 27, 13, 39, 19), or (33, 99, 49, 147, 73, 219, 109, 327, 163, 81, 243, 121, 363, 181, 543, 271, 135, 67). (End)


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 1..10001


PROG

(PARI) A135730(n)=local(c=0); while( n>17  n != 17 && n != 5 && n != 1, c++; if( n%2, n=3*n1, n>>=1)); c


CROSSREFS

Cf. A001281, A037084, A039500A039505, A135727A135729. A006370, A006577 (Collatz 3x+1 problem).
Cf. also A261671.
See A261673 and A261674 for records.
Sequence in context: A136715 A077116 A249507 * A188595 A144102 A195388
Adjacent sequences: A135727 A135728 A135729 * A135731 A135732 A135733


KEYWORD

easy,nonn


AUTHOR

M. F. Hasler, Nov 26 2007


STATUS

approved



