A061438 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).

25, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 36, 42, 34, 40, 32, 38, 30, 36, 28, 34, 26, 32, 24, 30, 34, 113, 28, 32, 111, 26, 30, 109, 24, 28, 107, 22, 26, 105, 36, 28, 24, 103, 34, 26, 22, 101, 32, 24, 20, 99, 30, 22, 24, 166, 97, 28, 20, 22, 164, 95, 26, 18, 20, 162, 93

Offset

1,1

Comments

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.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cino Hilliard, The x+1 conjecture

Example

x = 12: step 1: x = 12*14+13-12 = 169, step 2: x = 169/13 = 13, step 3: x = 13/13 = 1. Count = 3.

Mathematica

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 *)

Prog

(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" ") ) }

Crossrefs

Sequence in context: A171806 A038822 A375335 * A022981 A023467 A158501

Adjacent sequences: A061435 A061436 A061437 * A061439 A061440 A061441

Keyword

easy,nonn

Author

Cino Hilliard, Mar 29 2003

Status

approved