A346369 a(n) is the number of steps the Collatz trajectory of -n takes to reach a cycle, or -1 if no cycle is ever reached.

0, 0, 3, 1, 0, 4, 0, 2, 7, 0, 5, 5, 5, 0, 8, 3, 0, 8, 3, 0, 6, 6, 1, 6, 0, 6, 3, 1, 9, 9, 4, 4, 11, 0, 9, 9, 0, 4, 12, 1, 0, 7, 7, 7, 5, 2, 12, 7, 9, 0, 7, 7, 15, 4, 0, 2, 28, 10, 10, 10, 0, 5, 15, 5, 36, 12, 3, 0, 18, 10, 18, 10, 7, 0, 5, 5, 13, 13, 13, 2, 18

Offset

1,3

Comments

The analog of A139399 for negative starting values.

Is a(n) > -1 for all n?

Links

Table of n, a(n) for n=1..81.

Formula

a(n) = A224166(n) - A224254(n) = 1 + A224183(n) - A224254(n). - Pontus von Brömssen, Jul 24 2021

Example

For n = 5: The trajectory of -5 starts -5, -14, -7, -20, -10, -5, -14, -7, -20, -10, -5, ..., with -5 already being part of the cycle {-5, -14, -7, -20, -10}, so a(5) = 0.
For n = 6: The trajectory of -6 starts -6, -3, -8, -4, -2, -1, -2, -1, -2, -1, -2, ..., reaching a term of the cycle {-2, -1} after 4 steps, so a(6) = 4.

Prog

(PARI) a006370(n) = if(n%2==0, n/2, 3*n+1)
trajectory(n, terms) = my(v=[n]); while(#v < terms, v=concat(v, a006370(v[#v]))); v
a(n) = my(t, v=[]); for(k=1, oo, t=trajectory(-n, k); for(x=1, #t, if(x < #t && t[x]==t[#t], return(x-1))))

Crossrefs

Cf. A139399, A224166, A224183, A224254.

Sequence in context: A285123 A159059 A340262 * A127569 A117372 A127570

Adjacent sequences: A346366 A346367 A346368 * A346370 A346371 A346372

Keyword

sign

Author

Felix Fröhlich, Jul 14 2021

Status

approved