cp's OEIS Frontend

If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).

The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017

Only these numbers appear in A060445, which tabulates the "dropping time" of odd numbers. Note that all even numbers have a "dropping time" of 1.

a(n) is also the number of binary digits of 6^(n-1); for example, a(4)=8 since 6^(4-1)=216 in binary is 11011000, an 8-digit number. - Julio Cesar de la Yncera, Mar 28 2009

A positive integer (x) is an allowable value if and only if (x-1)/(1+log(2)/log(3)) - floor(x/(1+log(2)/log(3))) is not negative. - K. Spage, Oct 22 2009

Here the word "allowable" means that it is necessary for a sequence of iterates starting from odd value m to arrive at a value x = f^{floor(1+n+n*log(3)/log(2))}(m) < m, where n gives the number of odds in such a sequence including m, to have undergone precisely floor(1+n+n*log(3)/log(2)) iterations of f, where f(2*m)=m, f(2*m+1)=6*m+4. However, the formula for a(n+1) does not fully account for the order of odds and evens in such a sequence because it does not account for the effects of the "+1". Thus it is unknown whether it maximizes the value x for all values m. For example, fix m = 1 and the "+1" is enough to give the trivial cycle. So it is possible that for some m we have f^{floor(1+n+n*log(3)/log(2))}(m) >= m. - Jeffrey R. Goodwin, Aug 24 2011

The indices of the powers of 3 in A006899. - Ruud H.G. van Tol, Nov 02 2022

The dropping time is the number of Collatz iterations required to reach a lower number than starting value. Garner mentions these congruences. The first term in row n is A122442(n+1) for n > 1. The length of row n is A100982(n).

The triangle means:

numbers 0 (mod 2) and > 0 have dropping time 1;

numbers 1 (mod 4) and > 1 have dropping time 3;

numbers 3 (mod 16) have dropping time 6;

numbers 11, 23 (mod 32) have dropping time 8;

numbers 7, 15, 59 (mod 128) have dropping time 11;

numbers 39, 79, 95, 123, 175, 199, 219 (mod 256) have dropping time 13.

Theorem: a(n) can be evaluated using a directed rooted tree produced by a precise algorithm. Each node of this tree is given by a unique Diophantine equation whose only positive solutions are the integers with a finite stopping time. The algorithm generates (in a three step loop) the parity vectors which define the Diophantine equations. The two directions of the construction principle gives the tree a triangular form which extends ever more downwards with each column. There exist explicit arithmetic relationships between the parent and child vertices. As a consequence, a(n) can be generated algorithmically. The algorithm also generates A100982. - Mike Winkler, Sep 12 2017

Consider A126241, the sequence of dropping times in the Collatz iteration. Only zero and the numbers in A020914 can be dropping times. The dropping times in A126241 have a definite pattern. For example, 1 appears at positions n = 2 + 2*i, for i=0,1,2,3,... Similarly, 2 appears at positions n = 5 + 4*i; 4 appears at n = 3 + 16*i; 5 appears at n = 11 + {12,32}*i; and 7 appears at 7 + {8, 52, 128}*i. In general, if we let s=A020914(r) be the r-th possible stopping time, then A126241(n) = s for n = A122442(r) + T(r)*i, where T(r) is the r-th row of this triangle. The length of row n is A186009(n). The n-th row ends with 2^A020914(n).

The frequency of the r-th dropping time s=A020914(r) can be computed as A186009(r)/2^s. The first few frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, and 85/8192.

The term "stopping time" is sometimes used instead of "dropping time", but the former usually refers to A006666.

This sequence is closely related to A177789.