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

Author

Cino Hilliard, Mar 29 2003

Keywords

• easy
• nonn

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.

Examples

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

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000
• Cino Hilliard, The x+1 conjecture

Programs

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