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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. 7
 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. Pureferret, Longest known sequence of identical consecutive Collatz sequence lengths, Mathematics StackExchange, 2013. Carlos Rivera, Puzzle 847: Consecutive primes with the same Collatz length, The Prime Puzzles and Problems Connection. Carlos Rivera, Puzzle 851: Puzzle 847 revisited, The Prime Puzzles and Problems Connection. 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 June 21 08:18 EDT 2021. Contains 345358 sequences. (Running on oeis4.)