cp's OEIS Frontend

On the set of positive integers, the orbit of any number seems to end in the orbit of 1, of 5 or of 17. Writing n=1+q*2^p with q odd, it is easily seen that for p=0,1 and p>3, some iterations of the map lead to a strictly smaller number (for n>17). The cases p=2 and p=3 may give rise to bigger loops (depending on the form of q). See sequences A135727-A135729 for maxima of the orbits and corresponding record indices. - M. F. Hasler, Nov 29 2007

It appears that lim_{n->oo} a(n)/n = 5/2. - Benoit Cloitre, Jan 29 2006

Equivalent to the Collatz ('3n+1') problem for negative integers. - Dmitry Kamenetsky, Jan 12 2017

There are 327679 terms in this sequence which are less than 1000000. Based on this, I would suggest that the limit of a(n)/n is more likely to be 3 than 5/2. This is also a natural guess; there are three known periodic orbits for this recurrence. - David Speyer, Mar 25 2022

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 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.

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)

On the set of positive integers, the orbit of any number under A001281 seems to end in the orbit of 2, of 20 or of 272, which are the respective maxima of these cycles. Since any odd number increases under the map A001281, all elements of this sequence are even.

This gives indices n for which A135727(n) is larger than all preceding values of that sequence. As in A135727(n), we include the fixed point 0 in the domain of A001281. Obviously, many but not all entries are of the form 2^k+1 and not all of such numbers are in the sequence (e.g. 257, 1025, 2049 are missing). Is there a simple way of characterizing the exceptions?

This subsequence of A135728 gives indices n for which A135727(n)/n (ratio of maximal value to starting value) is larger than for all preceding indices. Obviously, we cannot consider the index n=0 here.