A061436 Number of steps for trajectory of n to reach 1 under the map that sends x -> x/3 if x mod 3 = 0, x -> x+3-(x mod 3) if x is not 0 mod 3 (for a 2nd time when n starts at 1).

2, 2, 1, 4, 4, 3, 3, 3, 2, 6, 6, 5, 6, 6, 5, 5, 5, 4, 5, 5, 4, 5, 5, 4, 4, 4, 3, 8, 8, 7, 8, 8, 7, 7, 7, 6, 8, 8, 7, 8, 8, 7, 7, 7, 6, 7, 7, 6, 7, 7, 6, 6, 6, 5, 7, 7, 6, 7, 7, 6, 6, 6, 5, 7, 7, 6, 7, 7, 6, 6, 6, 5, 6, 6, 5, 6, 6, 5, 5, 5, 4, 10, 10, 9, 10, 10, 9, 9, 9, 8, 10, 10, 9, 10, 10, 9, 9, 9, 8, 9, 9

Offset

1,1

Comments

This sequence is generated by the first PARI program below for m=3, p=1. 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=7 for n up to 100000. 100000 requires 157 steps to reach 1.

Links

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cino Hilliard, The x+1 conjecture

Example

x=1. step 1: x = 1+3-1 = 3; step 2: x = 3/3 = 1. Count: 2 steps.

Prog

(PARI) multxp2(n, m, p) = { print1(2" "); for(j=1, n, x=j; c=0; while(x>1, r = x%m; if(r==0, x=x/m, x=x*p+m-r); print1(x" "); ); ) }
(PARI) A061436(n) = if(1==n, 2, my(c=0); while(n>1, if(!(n%3), n = n/3, n += (3-(n%3))); c++); (c)); \\ Antti Karttunen, Apr 05 2022

Crossrefs

Sequence in context: A110664 A193922 A319534 * A214095 A213948 A136787

Adjacent sequences: A061433 A061434 A061435 * A061437 A061438 A061439

Keyword

easy,nonn

Author

Cino Hilliard, Mar 29 2003

Status

approved