cp's OEIS Frontend

Suppose that b and c are integers satisfying 1 < b < c. Let x = 1 + log_b(c) and y = 1 + log_c(b). Jointly rank all the numbers b^k for k>=0 and c^k for k>=1; then for n >= 0, the position of b^n is 1 + floor(n*y), and for n >=1, the position of c^n is 1+ floor(n*x).

These position sequences are closely related to the Beatty sequences given by floor(n*x) and floor(n*y).

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

Also, numbers k such that the first digit in the ternary expansion of 2^k is 1. - Mohammed Bouayoun (Mohammed.bouayoun(AT)sanef.com), Apr 24 2006

a(n) is the smallest integer such that n/a(n) < log_2(3). - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 31 2008

This sequence represents allowable values of the "dropping time" in the Collatz (3x+1) problem when iterated according to the function f(n) := n/2 if n is even, (3n+1)/2 otherwise, as tabulated in A126241. There is one exception, A126241(1), which has been set to zero by convention. - K. Spage, Oct 22 2009

An integer k is a term of A020914 if and only if floor(k*(1 + log(2)/log(3))) - abs(k-1)*(1 + log(2)/log(3)) - 1 >= 0. - K. Spage, Oct 22 2009

Also smallest k such that ceiling(2^k / 3^n) = 2. - Michel Lagneau, Jan 31 2012

For n > 0, first differences of A022330. - Michel Marcus, Oct 03 2013

Also the number of powers of two less than or equal to 3^n. - Robert G. Wilson v, May 25 2014

Except for 1, A020914 is the complement of A054414 and therefore these two form a pair of Beatty sequences. - Robert G. Wilson v, May 25 2014

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

Original name was: Smallest exponent of 2 which gives a power of 2 which is equal to or bigger than (3/2)^n, n = 0,1,... .

Stacking perfect fifths (the frequency ratio of a fifth is 3/2) this sequence determines into which octave the n-th fifth falls. For example, the third fifth, (3/2)^3, falls into the second octave, which means that it lies in the interval [2^1,2^2)=[2,4). The k-th octave comprises ratios in the interval [2^(k-1),2^k), k=1,2,...

Related to the initial number of sequential even terms in an "ideal" sequence under iteration of the 3x+1 Problem on a positive odd value m, where the piecewise function f is given by f(2*m)=m, f(2*m+1)=6*m+4, to ensure f^A122437(n) (m) < m, where n > 1 is the number of odds in the sequence (including m) and floor(1+n*(log(3)/log(2))) is the number of evens. An "ideal" sequence minimizes the effects of f(2*m+1) by following a certain order of even or odd terms along with the rules of the function. A representation of such sequences in terms of parity sequences for values n >= 2 follows:

n=2, (o,e,e,o,e,e)

n=3, (o,e,e,o,e,o,e,e)

n=4, (o,e,e,e,o,e,o,e,o,e,e)

n=5, (o,e,e,e,o,e,o,e,o,e,o,e,e)

n=6, (o,e,e,e,e,o,e,o,e,o,e,o,e,o,e,e)

n=7, (o,e,e,e,e,e,o,e,o,e,o,e,o,e,o,e,o,e,e)

The pattern is clear, and the formula for the initial number of sequential even terms in each sequence is given by a(n) = floor(1+n*(log(3)/log(2)))-n for n > 1, where the sum of the number of even and odd terms is given by A122437(n) for n > 1. Of course, most values m do not have sequences following this pattern of iteration under f. Also, the reason for placing an extra even term at the end of such sequences is to mitigate to some degree the effects of the possibility that the last odd term is only "slightly" larger than m, i.e., (3*m+1)/4 < m for all m > 1. - Jeffrey R. Goodwin, Aug 25 2011

a(n) gives the position in n-th row of A227048 where (3^n - 2^n) occurs:

A227048(n,a(n)) = A001047(n). - Reinhard Zumkeller, Jun 30 2013

Differs from A005378 at indices n = 0,17,20,22,25,27,29,30,... - M. F. Hasler, Jun 29 2014

Represents increments between successive terms of allowable dropping times in the Collatz (3x+1) problem. That is, a(n) = A020914(n+1) - A020914(n). - K. Spage, Oct 23 2009

The joint ranking is for j >= 1 and k >= 1, so that the sets {2^j} and {3^k} are disjoint.

A unimodal Collatz glide sequence is successive rises x -> (3x+1)/2 followed by successive falls x -> x/2 until dropping below its starting x.

After n increases, there are ceiling(n*log(3)/log(2) - n) decreases to drop below the initial value.