|
%I #17 Jun 14 2017 00:42:09
%S 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,
%T 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,
%U 11,11,7,7,19,6,37,20,20,2,19,14,19,14,15,15,9,9,14,14,14,4
%N Number of steps to reach the minimum of the final cycle under iterations of the map A001281: x->3x-1 if x odd, x/2 otherwise.
%C 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.
%C From _N. J. A. Sloane_, Sep 04 2015: (Start)
%C The same sequence arises as follows: Start at 2n-1 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.
%C 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)
%H N. J. A. Sloane, <a href="/A135730/b135730.txt">Table of n, a(n) for n = 1..10001</a>
%o (PARI) A135730(n)=local(c=0);while( n>17 || n != 17 && n != 5 && n != 1, c++; if( n%2, n=3*n-1,n>>=1));c
%Y Cf. A001281, A037084, A039500-A039505, A135727-A135729. A006370, A006577 (Collatz 3x+1 problem).
%Y Cf. also A261671.
%Y See A261673 and A261674 for records.
%K easy,nonn
%O 1,3
%A _M. F. Hasler_, Nov 26 2007
|
