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A277109 Starting from 2^n+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem. 5
1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 30, 1, 1, 3, 7, 15, 26, 26, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 7, 15, 1, 1, 3, 7, 15, 26, 1, 3, 7, 15, 31, 63, 26, 30, 26, 26, 26, 26, 30, 46, 26, 26, 26, 26, 1, 1, 1, 3, 7, 15, 1, 3, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,7

COMMENTS

a(500) was found by Guo-Gang Gao (see links).

Interestingly, this sequence has many sets of consecutive terms that are increasing powers of 2 minus 1. For example: a(291) to a(307), a(447) to a(467), and a(603) to a(625). It is not clear why this is the case.

The largest value known in this sequence is a(1812) = 2^26-1 = 67108863.

Conjecture: If a(n) = 2^k - 1 for some k > 1, then a(n-1) = 2^(k-1) - 1. Conjecture holds for n <= 1812.

From Hartmut F. W. Hoft, Aug 16 2018: (Start)

The conjecture is true. Let the lengths of the Collatz runs equal q for all numbers 2^n + 1, 2^n + 2, 2^n + 3, 2^n + 4, ..., 2^n + 2^k - 2, 2^n + 2^k - 1. Then dividing the 2^(k-1) - 1 even numbers by two gives rise to the sequence 2^(n-1) + 1, 2^(n-1) + 2, ..., 2^(n-1) + 2^(k-1) - 1 of numbers for which the lengths of the Collatz runs equals q-1. Furthermore, let the length of the Collatz run of 2^n + 2^k be r != q then the length of the Collatz run of 2^(n-1) + 2^(k-1) is r-1 != q-1, i.e., a(n-1) = 2^(k-1) - 1.

Conjecture: Let a(k), ..., a(k+m), m >= 0, be a subsequence of this sequence such that a(k)=a(k+m+1)=1 and a(k+i) > 1, 1 <= i <= m. Then the lengths of the Collatz runs of a(k+i), 0 <= i <= m, increase by 1. In addition, there is an initial segment of increasing numbers a(k), ..., a(k+j), for some 0 <= j <= m, in each such subsequence having the form 2^i - 1, 0 < i <= j. (End)

LINKS

Dmitry Kamenetsky, Table of n, a(n) for n = 0..1812

Dmitry Kamenetsky, C program to compute the sequence

Guo-Gang Gao, On consecutive numbers of the same height in the Collatz problem, Discrete Mathematics, Volume 112, pages 261-267, 1993.

Carlos Rivera, Puzzle 847: Consecutive primes with the same Collatz length

Carlos Rivera, Puzzle 851: Puzzle 847 revisited

EXAMPLE

a(6) = 3, because 2^6+1, 2^6+2 and 2^6+3 all take 27 steps to reach 1.

From Hartmut F. W. Hoft, Aug 16 2018: (Start)

Two examples for the conjecture (L(n) denotes the length of the Collatz run):

n      a(n)    L(n)          n      a(n)    L(n)

64     26      483           20      1       72

------------------           ------------------

65      1      559           21      1      166

66      3      560           22      3      167

67      7      561           23      7      168

68     15      562           24     15      169

69     31      563           ------------------

70     63      564           25     26      170

------------------           26     26      171

71     26      565           ------------------

72     30      566           27      1      247

73     26      567

74     26      568

75     26      569

76     26      570

77     30      571

78     46      572

79     26      573

80     26      574

81     26      575

82     26      576

------------------

83      1      626

The "power of 2 minus 1" initial section of any such subsequence of a(n) is always increasing. However, there is no apparent ordering in the second section when that is present. (End)

MATHEMATICA

f[n_] := Length[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # != 1 &]] - 1; Table[k = 1; While[f[2^n + k] == f[2^n + k + 1], k++]; k, {n, 120}] (* Michael De Vlieger, Oct 03 2016 *)

PROG

nbsteps(n)= s=n; c=0; while(s>1, s=if(s%2, 3*s+1, s/2); c++); c;

a(n) = {my(ns = 2^n+1); my(nbs = nbsteps(ns)); while(nbsteps(ns+1) == nbs, ns++); ns - 2^n; } \\ Michel Marcus, Oct 30 2016

CROSSREFS

Cf. A006577, A078441, A179118, A277684.

Sequence in context: A062174 A154754 A102368 * A063062 A066056 A153284

Adjacent sequences:  A277106 A277107 A277108 * A277110 A277111 A277112

KEYWORD

nonn

AUTHOR

Dmitry Kamenetsky, Sep 30 2016

STATUS

approved

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Last modified April 20 06:19 EDT 2019. Contains 322294 sequences. (Running on oeis4.)