%I #41 Mar 22 2021 03:41:34
%S 0,6,3,11,3,8,3,11,3,6,3,8,3,96,3,91,3,6,3,13,3,8,3,88,3,6,3,8,3,11,3,
%T 88,3,6,3,83,3,8,3,13,3,6,3,8,3,73,3,13,3,6,3,68,3,8,3,50,3,6,3,8,3,
%U 13,3,24,3,6,3,11,3,8,3,11,3,6,3,8,3,65,3,34,3,6,3,47,3,8,3,13,3,6,3,8,3
%N "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).
%C If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).
%C The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - _Juhani Heino_, Jul 21 2017
%H T. D. Noe, <a href="/A060445/b060445.txt">Table of n, a(n) for n = 0..10000</a>
%H Jason Holt, <a href="/A006577/a006577_1B.png">Plot of first 1 billion terms</a>, log scale on x axis
%H Jason Holt, <a href="/A006577/a006577_10B.png">Plot of first 10 billion terms</a>, log scale on x axis
%H Eric Roosendaal, <a href="http://www.ericr.nl/wondrous/index.html">On the 3x + 1 problem</a>
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%e 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.
%t nxt[n_]:=If[OddQ[n],3n+1,n/2]; Join[{0},Table[Length[NestWhileList[nxt, n,#>=n&]]-1, {n,3,191,2}]] (* _Harvey P. Dale_, Apr 23 2011 *)
%o (Haskell)
%o a060445 0 = 0
%o a060445 n = length $ takeWhile (>= n') $ a070165_row n'
%o where n' = 2 * n + 1
%o -- _Reinhard Zumkeller_, Mar 11 2013
%o (Python)
%o def a(n):
%o if n<1: return 0
%o n=2*n + 1
%o N=n
%o x=0
%o while True:
%o if n%2==0: n//=2
%o else: n = 3*n + 1
%o x+=1
%o if n<N: break
%o return x
%o print([a(n) for n in range(101)]) # _Indranil Ghosh_, Apr 22 2017
%Y A060565 gives the first lower number that is reached. Cf. A060412-A060415, A217934.
%Y See A074473, A102419 for other versions of this sequence.
%Y Cf. A122437 (allowable dropping times), A122442 (least k having dropping time A122437(n)).
%Y Cf. A070165.
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_, Apr 07 2001
%E More terms from _Jason Earls_, Apr 08 2001 and from _Michel ten Voorde_ Apr 09 2001
%E Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001