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A248037
Numbers n such that the ratio of tripling steps to halving steps in the Collatz (3x+1) trajectory of n is greater than all previous ratios.
3
2, 3, 7, 9, 27, 230631, 626331, 837799, 1723519, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231
OFFSET
1,1
COMMENTS
Equivalently, numbers n such that A006667(n)/A064433(n) > A006667(m)/A064433(m) for all 0 < m < n.
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)
From Jon E. Schoenfield, Nov 21 2015: (Start)
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 b-file 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
PROG
(PARI)
Tratio(n)=c=0; d=0; while(n!=1, if(n==Mod(0, 2), n=n/2; d++); if(n==Mod(1, 2)&&n!=1, n=3*n+1; c++)); c/d
print1(2, ", "); n=2; p=Tratio(2); while(n, t=Tratio(n+1); if(p>=t, n+=2); if(p<t, print1(n+1, ", "); p=t; n+=2))
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Derek Orr, Sep 29 2014
EXTENSIONS
Corrected and extended by Sean A. Irvine, Derek Orr, and David Consiglio, Jr., Nov 23 2015
a(16) from David Consiglio, Jr. and Sean A. Irvine, Nov 26 2015
a(17) added by Dmitry Kamenetsky, Sep 23 2016
STATUS
approved