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A336912

Number of steps to reach 1 in the 'x^3+1' problem (a variation of the Collatz problem), or -1 if 1 is never reached.

0, 1, 9, 2, 7, 2, 5, 2, 10, 10, 5, 10, 5, 10, 8, 3, 5, 3, 13, 3, 13, 3, 13, 3, 16, 8, 8, 8, 13, 8, 8, 8, 11, 8, 11, 3, 26, 3, 21, 3, 6, 3, 8, 3, 8, 3, 8, 3, 16, 6, 16, 6, 16, 6, 16, 6, 6, 6, 16, 6, 16, 6, 6, 3, 6, 3, 16, 3, 21, 3, 6, 3, 11, 3, 11, 3, 29, 3

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OFFSET

1,3

COMMENTS

The x^3+1 map, which is a variation of the 3x+1 (Collatz) map, is defined for x >= 0 as follows: if x is odd, then map x to x^3+1; otherwise, map x to floor(sqrt(x)).

It seems that all x^3+1 trajectories reach 1; this has been verified up to 10^10.

LINKS

Table of n, a(n) for n=1..78.

Wikipedia, Collatz conjecture.

EXAMPLE

For n = 3, a(3) = 9, because there are 9 steps from 3 to 1 in the following trajectory for 3: 3, 28, 5, 126, 11, 1332, 36, 6, 2, 1.

For n = 4, a(4) = 2, because there are 2 steps from 4 to 1 in the following trajectory for 4: 4, 2, 1.

PROG

(Python)

from math import isqrt

def a(n):

count = 0