A344276 Number of halving and tripling steps to reach 3 in the '3x+3' problem, or -1 if 3 is never reached.

2, 3, 0, 4, 9, 1, 4, 5, 7, 10, 10, 2, 18, 5, 5, 6, 21, 8, 8, 11, 16, 11, 11, 3, 11, 19, 19, 6, 19, 6, 6, 7, 14, 22, 22, 9, 22, 9, 9, 12, 9, 17, 17, 12, 17, 12, 12, 4, 25, 12, 12, 20, 113, 20, 20, 7, 20, 20, 20, 7, 108, 7, 7, 8, 28, 15, 15, 23, 15, 23, 23, 10

Offset: 1

Author

Matthias van Sligtenhorst, May 13 2021

Keywords

• nonn

Comments

The '3x+3' problem is a slight variation of the Collatz problem. If n is even, divide it by 2, if n is odd, multiply by 3 and add 3. The number of steps to reach 3 are given, which may be the endpoint for all n (empirical observation).

From Jon E. Schoenfield, May 14 2021: (Start)

It seems that the average number of steps among the '3x+3' trajectories for n in 1..3m is close to the average number of steps in the '3x+1' trajectories for n in 1..m:

.

m (Sum_{n=1..m} a(n))/m (Sum_{n=1..3m} c(n))/3m

---- --------------------- -----------------------

10^1 6.7000000000 8.6666666667

10^2 31.4200000000 32.1466666667

10^3 59.5420000000 58.9020000000

10^4 84.9666000000 84.6180333333

10^5 107.5384000000 107.6915966667

where c(n) = A006577(n) is the number of steps in the '3x+1' trajectory of n.

Perhaps a good way to explain this result is that, other than the values connected by the string of consecutive divide-by-2 steps at the beginning of the trajectory of an even number not divisible by 3, every value in every '3x+3' trajectory is a multiple of 3, so within any given interval, there are only about 1/3 as many values available for inclusion in '3x+3' trajectories as there are in '3x+1' trajectories. (End)

Examples

			a(1) = 2, with the trajectory 1 -> 6 -> 3.
a(5) = 9, with the trajectory 5 -> 18 -> 9 -> 30 -> 15 -> 48 -> 24 -> 12 -> 6 -> 3.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A067896 (trajectory of 41).

Cf. A006577 (3x+1 steps).

Programs

Maple

a:= proc(n) a(n):= 1+a(`if`(n::odd, 3*n+3, n/2)) end: a(3):=0:
seq(a(n), n=1..100);  # Alois P. Heinz, May 14 2021

Mathematica

If[#!=3,#0@If[OddQ@#,3#+3,#/2]+1,0]&/@Range@100 (* Giorgos Kalogeropoulos, May 14 2021 *)

PARI

a(n) = for (k=0, oo, if (n==3, return (k), n%2==0, n=n/2, n=3*n+3)) \\ Rémy Sigrist, Jun 13 2021

Formula

a(3) = 0; for all other n > 0, if n is even, a(n) = a(n/2) + 1; if n is odd, a(n) = a(3n+3) + 1.