A006667(n) is the number of tripling steps in the Collatz (3x+1) problem and A064433(n) is the number of halving steps in the Collatz (3x+1) problem.
It is crucial to make A006667(n) the numerator as it can be zero when n = 2^k for some k > 0.
a(n) is odd for all n > 1.
The corresponding ratios are:
0.0000000000000000000000000000... (2)
0.4000000000000000000000000000... (3)
0.4545454545454545454545454545... (7)
0.4615384615384615384615384615... (9)
0.5857142857142857142857142857... (27)
0.5899280575539568345323741007... (230631)
0.5924764890282131661442006269... (626331)
0.5927051671732522796352583586... (837799)
0.5931232091690544412607449856... (1723519)
0.5935828877005347593582887700... (3732423)
0.5937500000000000000000000000... (5649499)
0.5961538461538461538461538461... (6649279)
0.5967365967365967365967365967... (8400511)
0.6030405405405405405405405405... (63728127)
0.6035196687370600414078674948... (3743559068799)
If we define a "tripling step" to also include a "halving step" afterwards (since 3*n+1 converts an odd number n into an even number, so a halving step will always follow), the ratios are still maximum at the a(n) values. However, the ratios themselves are different. The corresponding ratios in this case are:
0.000000000000000000000000000... (2)
0.666666666666666666666666666... (3)
0.833333333333333333333333333... (7)
0.857142857142857142857142857... (9)
1.413793103448275862068965517... (27)
1.438596491228070175438596491... (230631)
1.453846153846153846153846153... (626331)
1.455223880597014925373134328... (837799)
1.457746478873239436619718309... (1723519)
1.460526315789473684210526315... (3732423)
1.461538461538461538461538461... (5649499)
1.476190476190476190476190476... (6649279)
1.479768786127167630057803468... (8400511)
1.519148936170212765957446808... (63728127)
1.656946826758147512864493997... (3743559068799)
Let T and H be the number of tripling steps and halving steps, respectively, in the entire Collatz (3x+1) trajectory of a number n. Since each tripling step increases the value by a factor greater than 3, and each halving step decreases it by a factor of exactly 2, we have n * 3^T / 2^H < 1, from which T/H < log(2)/log(3)  log_3(n)/H, so the ratio T/H cannot exceed log(2)/log(3) = 0.6309297535...
It seems likely that the present sequence is a subsequence of A006877 (which consists of values n whose trajectories are of record length). Taking as values of n the terms from the bfile for A006877, and generating their trajectories to obtain the corresponding values of H(n), it does not seem clear whether log_3(n)/H(n) is converging toward zero or toward some positive limit, perhaps around 0.020 (which would mean T/H < log(2)/log(3)  0.020, i.e., T/H < 0.611).
The known terms n in A006877 at which log_3(n)/H(n) reaches a record low coincide almost exactly with the known terms of this sequence, the only exception thus far being a(11) = A006877(52) = 5649499, at which log_3(n)/H(n) is only slightly larger than it is at a(10) = A006877(51) = 3732423 (0.03685302 vs. 0.03682956). Given the values of log_3(n)/H(n) for the remaining known terms in A006877, it seems likely that
a(16) = A006877(110) = 100759293214567
and that a(17) exceeds A006877(130), which is 46785696846401151.
(End)
Note that a(17)=104899295810901231 has now been found by Eric Roosendaal's distributed project (see link below).  Dmitry Kamenetsky, Sep 23 2016
For n>=14, a(n) must be 7, 15, 27, or 31 (mod 32). This is because all other values provably have a ratio of tripling to halving steps of less than 0.6 (see program by Irvine and Consiglio Jr.).  Dmitry Kamenetsky, Sep 24 2016
