A265287 - OEIS

%I #9 Dec 13 2015 07:57:31

%S 0,1,5,2,4,6,6,3,14,5,5,7,13,7,7,4,18,15,12,6,9,6,20,8,17,14,11,8,11,

%T 8,8,5,19,19,19,16,19,13,10,7,13,10,13,7,7,21,18,9,21,18,18,15,18,12,

%U 12,9,30,12,12,9,12,9,9,6,23,20,17,20,23,20,23,17,17

%N Number of steps needed to reach 1 in the "sqrt(2)*x + 1" problem.

%C The sqrt(2)*x + 1 problem is as follows: start with a number x. If x is even, divide it by 2, otherwise multiply it by sqrt(2) and add 1, and then take the integer part.

%C Conjecture: the trajectory reaches 1 for all n.

%C Generalization:

%C If we consider the "sqrt(3)*x + 1" problem (see A264789), we observe three possible behaviors for such trajectories when n>0:

%C (i) The trajectory reaches 1 (and enters the "trivial" cycle 2-1-2-1-2...).

%C (ii) Cyclic trajectory. The trajectory becomes periodic and the period does not contain a 1.

%C (iii) The trajectory is divergent (I conjecture that this cannot occur).

%C If we consider the "sqrt(q)*x+1" problem with q>3 but different from 9, we observe divergent trajectories.

%H Michel Lagneau, <a href="/A265287/b265287.txt">Table of n, a(n) for n = 1..10000</a>

%e a(7) = 6 because 7 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1 with 6 iterations where:

%e 10 = floor(7*sqrt(2)+1);

%e   5 = 10/2;

%e   8 = floor(5*sqrt(2)+1);

%e   4 = 8/2;

%e   2 = 4/2 and 1 = 2/2 is the end of the cycle.

%t f[n_]:=Module[{a=n,k=0},While[a!=1,k++;If[EvenQ[a],a=a/2,a=Floor[a*Sqrt[2]+1]]];k];Table[f[n],{n,100}]

%Y Cf. A006577,  A264789.

%K nonn

%O 1,3

%A _Michel Lagneau_, Dec 06 2015