

A208981


Number of iterations of the Collatz recursion required to reach a power of 2.


2



0, 0, 3, 0, 1, 4, 12, 0, 15, 2, 10, 5, 5, 13, 13, 0, 8, 16, 16, 3, 1, 11, 11, 6, 19, 6, 107, 14, 14, 14, 102, 0, 22, 9, 9, 17, 17, 17, 30, 4, 105, 2, 25, 12, 12, 12, 100, 7, 20, 20, 20, 7, 7, 108, 108, 15, 28, 15, 28, 15, 15, 103, 103, 0, 23, 23, 23, 10, 10, 10
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OFFSET

1,3


COMMENTS

The statement that all paths must eventually reach a power of 2 is equivalent to the Collatz conjecture.
A006577(n)  a(n) gives the exponent for the first power of 2 reached in the Collatz trajectory of n.  Alonso del Arte, Mar 05 2012


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
John Smith, Collatz sequence, PlanetMath.
Eric Weisstein's World of Mathematics, Collatz problem.
Index entries for sequences related to 3x+1 (or Collatz) problem


FORMULA

For x>0 an integer, define f_0(x)=x, and for r=1,2,..., f_r(x)=f_{r1}(x)/2 if f_{r1}(x) is even, else f_r(x)=3*f_{r1}(x)+1. Then a(n) = min(k such that f_k(n) is equal to a power of 2).


EXAMPLE

a(7) = 12 because the Collatz trajectory for 7 is 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... which reached 16 = 2^4 in 12 steps.


MATHEMATICA

Collatz[n_?OddQ] := 3*n + 1; Collatz[n_?EvenQ] := n/2; Table[1 + Length[NestWhileList[Collatz, n, Not[IntegerQ[Log[2, #]]] &]], {n, 50}] (* Alonso del Arte, Mar 04 2012 *)


PROG

(Haskell)
a208981 = length . takeWhile ((== 0) . a209229) . a070165_row
 Reinhard Zumkeller, Jan 02 2013
(PARI) ispow2(n)=n>>=valuation(n, 2); n==1
a(n)=my(s); while(!ispow2(n), n=if(n%2, 3*n+1, n/2); s++); s \\ Charles R Greathouse IV, Jul 31 2016


CROSSREFS

Cf. A006577 (and references therein).
Cf. A209229, A070165, A010120.
Sequence in context: A213191 A079520 A229001 * A261158 A207543 A191532
Adjacent sequences: A208978 A208979 A208980 * A208982 A208983 A208984


KEYWORD

nonn,nice,look


AUTHOR

L. Edson Jeffery, Mar 04 2012


STATUS

approved



