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%I #28 Mar 22 2020 10:12:54
%S 1,2,7,2,4,2,12,2,4,2,9,2,4,2,12,2,4,2,7,2,4,2,9,2,4,2,97,2,4,2,92,2,
%T 4,2,7,2,4,2,14,2,4,2,9,2,4,2,89,2,4,2,7,2,4,2,9,2,4,2,12,2,4,2,89,2,
%U 4,2,7,2,4,2,84,2,4,2,9,2,4,2,14,2,4,2,7,2,4,2,9,2,4,2,74,2,4,2,14,2,4,2,7
%N Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.
%C Here we call the starting value iteration number 1, although usually the count is started at 0, which would subtract 1 from the values for n >= 2 - see A060445, A102419.
%H N. J. A. Sloane, <a href="/A074473/b074473.txt">Table of n, a(n) for n = 1..10000</a>
%e n=2k: then a(2k)=2 because the second iterate is k<n=2k, the first iterate below 2k; n=4k+1, k>1: the list = {4k+1, 12k+4, 6k+2, 3k+1, ...} i.e. the 4th term is always the first below initial value, so a(4k+1)=4;
%e n=15: the list={15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} and 12th term is first sinks below iv=15, so a(15)=12; relatively larger values occur at n=4k+3.
%e n=3: the list is {3, 10, 5, 16, 8, 4, 2, 1, ..}, the 7th term is 2, which is the first smaller than 3, so a(3)=7.
%t nextx[x_Integer] := If[OddQ@x, 3x + 1, x/2]; f[1] = 1; f[n_] := Length@ NestWhileList[nextx, n, # >= n &]; Array[f, 83] (* Bobby R. Treat (drbob(at)bigfoot.com), Sep 16 2006 *)
%o (Python)
%o def a(n):
%o if n<3: return n
%o N=n
%o x=1
%o while True:
%o if n%2==0: n/=2
%o else: n = 3*n + 1
%o x+=1
%o if n<N: return x
%o [a(n) for n in range(1, 101)] # _Indranil Ghosh_, Apr 15 2017
%Y Cf. A006370, A075476, A075477, A075478, A075479, A075480, A075481, A075482, A075483, A060445, A060412, A217934.
%Y Equals A102419(n)+1.
%K nonn
%O 1,2
%A _Labos Elemer_, Sep 19 2002
%E Edited by _N. J. A. Sloane_, Sep 15 2006
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