OFFSET
1,1
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)
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
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.
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Matthias van Sligtenhorst, May 13 2021
STATUS
approved