%I #12 Aug 11 2014 22:45:20
%S 25,23,21,19,17,15,13,11,9,7,5,3,1,36,42,34,40,32,38,30,36,28,34,26,
%T 32,24,30,34,113,28,32,111,26,30,109,24,28,107,22,26,105,36,28,24,103,
%U 34,26,22,101,32,24,20,99,30,22,24,166,97,28,20,22,164,95,26,18,20,162,93
%N Number of steps for trajectory of n to reach 1 under the map that sends x -> x/13 if x mod 13 = 0, x -> 14x+13-(x mod 13) if x is not 0 mod 13 (for a 2nd time when n starts at 1).
%C This sequence is generated by the program below for m=13,p=14. Other values of m and p also converge but not necessarily to 1. For m =2 and p=1 we have the count of steps for the x+1 problem. m=prime and p=m+1 usually converge to 1 but break down for certain values of n. E.g. 17 locks at n=34, 23 at n=49, 29 at n=91. I verified m=13 for n up to 100000. 100000 requires 100 steps to reach 1.
%H Harvey P. Dale, <a href="/A061438/b061438.txt">Table of n, a(n) for n = 1..1000</a>
%H Cino Hilliard, <a href="http://groups.msn.com/BC2LCC/page.msnw?fc_p=%2Fkx%2Bp%20problems&fc_a=0">The x+1 conjecture</a>
%e x = 12: step 1: x = 12*14+13-12 = 169, step 2: x = 169/13 = 13, step 3: x = 13/13 = 1. Count = 3.
%t Join[{25},Table[Length[NestWhileList[If[Divisible[#,13],#/13,14#+13-Mod[#,13]]&,n,#!=1&]],{n,2,70}]-1] (* _Harvey P. Dale_, Mar 14 2012 *)
%o (PARI) countxp2(n,m,p) = { c=1; x=1; x=x*p+m-1; while(x>1, r = x%m; if(r==0,x=x/m,x=x*p+m-r); c++; ); print1(c" "); for(j=2,n, x=j; c=0; while(x>1, r = x%m; if(r==0,x=x/m,x=x*p+m-r); c++; \ print1(x" "); ); print1(c" ") ) }
%K easy,nonn
%O 1,1
%A _Cino Hilliard_, Mar 29 2003