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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. 1
 2, 3, 7, 9, 27, 230631, 626331, 837799, 1723519, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231 (list; graph; refs; listen; history; text; internal format)
 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). 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 LINKS Sean A. Irvine and David Consiglio, Jr., C program Eric Roosendaal, Completeness and Gamma records 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

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