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A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1. 13

%I

%S 0,2,1,4,1,3,1,4,1,2,1,3,1,37,1,35,1,2,1,5,1,3,1,34,1,2,1,3,1,4,1,34,

%T 1,2,1,32,1,3,1,5,1,2,1,3,1,28,1,5,1,2,1,26,1,3,1,19,1,2,1,3,1,5,1,9,

%U 1,2,1,4,1,3,1,4,1,2,1,3,1,25,1,13,1,2,1,18,1,3,1,5,1,2,1,3,1,4,1,8,1,2,1,5

%N "Dropping time" of the reduced Collatz iteration starting with 2n+1.

%C We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: sum_{k>0} k*N(k)/P(k) = 3.492651852186...

%D Victor Klee and Stan Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, Mathematical Association of America (1991) pp. 225-229, 308-309. [called on p. 225 stopping time for 2n+1 and the function C(2*n+1) = A075677(n+1), n >= 0. - _Wolfdieter Lang_, Feb 20 2019]

%H T. D. Noe, <a href="/A122458/b122458.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) is the least k for which fr^[k](n) < 2*n + 1, for n >= 1 and k >= 1, where fr(n) = A075677(n+1) = A000265(3*n+2). No k satisfies this for n = 0: a(0) := 0 by convention. The dropping time a(n) is finite, for n >= 1, if the Collatz conjecture is true. - _Wolfdieter Lang_, Feb 20 2019

%e a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.

%e n = 13: the fr trajectory for 2*13+1 = 27 is 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1 with 41 terms (without 27), hence fr^[37] = 23 < 27 and a(13) = 37. - _Wolfdieter Lang_, Feb 20 2019

%t nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n],{n,1,301,2}]

%Y Cf. A000265, A060445, A075677 (one step of the reduced Collatz iteration), A075680.

%K nonn

%O 0,2

%A _T. D. Noe_, Sep 08 2006

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