A304030 a(n) is the number of steps at which the Collatz ('3x+1') trajectory of n crosses its initial value, or -1 if the number of crossings is infinite.

0, 0, 1, 0, 1, 4, 3, 0, 5, 2, 3, 2, 3, 8, 3, 0, 3, 10, 7, 0, 1, 6, 1, 0, 9, 2, 3, 6, 7, 4, 3, 0, 13, 4, 1, 4, 5, 8, 9, 0, 7, 2, 5, 2, 3, 4, 5, 0, 5, 8, 5, 0, 1, 14, 9, 0, 7, 2, 3, 6, 7, 12, 9, 0, 5, 6, 3, 0, 1, 4, 7, 0, 13, 2, 1, 2, 3, 8, 3, 0, 7, 14, 11, 0, 1, 8, 3, 0, 3, 2, 7, 4, 5, 12, 9, 0, 19, 4, 1, 0

Offset: 1

Author

Jon E. Schoenfield, May 04 2018

Keywords

• nonn

Comments

Some treatments of the Collatz conjecture view trajectories as starting to cycle when they reach 1, continuing with 4, 2, 1, 4, 2, 1, ..., while others view trajectories as terminating as soon as 1 is reached; this sequence treats trajectories as terminating at 1.

If the Collatz conjecture is true, then for n > 1, a(n) == n (mod 2).

If there exists any number n whose Collatz trajectory enters a cycle that includes values above and below n, then the number of crossings would be infinite. If the Collatz conjecture is true, then there exists no such value of n.

If a(k) = 0, then a(2^j * k) = 0, for j>0. Therefore the primitives are 1, 20, 24, 52, 56, 68, 72, 84, 88, 100, 116, ..., . - Robert G. Wilson v, May 19 2018

Examples

			The Collatz trajectory of 6 crosses its initial value (6) a total of 4 times, so a(6) = 4:
.
                           16
                           / \
                          /   \
              10         /     \
              / \       /       \
             /   \     /         8
  6---------*-----*---*-----------*--------
   \       /       \ /             \
    \     /         5               \
     \   /                           \
      \ /                             4
       3                               \
                                        ...
(Each "*" represents a crossing.)
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370, A070165, A349325.

Programs

Mathematica

Collatz[n_] := NestWhileList[ If[ OddQ@#, 3# +1, #/2] &, n, # > 1 &]; f[n_] := Block[{x = Length[ SplitBy[ Collatz@ n, # < n +1 &]] - 1}, If[ OddQ@ n && n > 1, x - 1, x]]; Array[f, 100] (* Robert G. Wilson v, May 05 2018 *)

Python

def A304030(n):
    prevc = c = n
    h = 0
    while c > 1:
        if c % 2:
            c = 3*c+1
            if prevc < n and c > n: h += 1
        else:
            c //= 2
            if prevc > n and c < n: h += 1
        prevc = c
    return h
print([A304030(n) for n in range(1, 100)]) # Paolo Xausa, Feb 22 2022