A349325 Number of times the Collatz plot started at n crosses the y = n line, or -1 if the number of crossings is infinite.

1, 1, 2, 1, 2, 3, 4, 1, 4, 1, 2, 1, 2, 5, 2, 1, 4, 5, 6, 1, 2, 3, 2, 1, 6, 1, 4, 1, 2, 3, 4, 1, 6, 1, 2, 1, 2, 3, 4, 1, 6, 1, 4, 1, 2, 3, 4, 1, 4, 1, 2, 1, 2, 7, 6, 1, 6, 1, 2, 3, 4, 7, 6, 1, 4, 1, 2, 1, 2, 3, 6, 1, 10, 1, 2, 1, 2, 5, 2, 1, 4, 5, 6, 1, 2, 3, 2

Offset

1,3

Comments

The plots considered are the trajectories from n to 1 of the 3x+1 function, denoted by T(x) in the literature, defined as T(x) = (3x+1)/2 if x is odd, T(x) = x/2 if x is even (A014682).

The starting value of the trajectory is considered a crossing.

A similar sequence for the "standard" Collatz function (A006370) is A304030.

Conjecture: every positive integer appears in the sequence infinitely many times.

References

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010.

Links

Table of n, a(n) for n=1..87.

J. C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.

Index entries for sequences related to 3x+1 (or Collatz) problem

Formula

a(2^k) = 1, for integers k >= 0.

a(A166245(m)) = 1 for m>=1. - Michel Marcus, Nov 16 2021

Example

The trajectory starting at 7 is 7 -> 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1, so the Collatz plot crosses the y = 7 line at the beginning, from 10 to 5, from 5 to 8 and from 8 to 4, for a total of 4 times. a(7) is therefore 4.

Mathematica

nterms=100; Table[h=1; prevc=c=n; While[c>1, If[OddQ[c], c=(3c+1)/2; If[prevc<n&&c>n, h++], c/=2^IntegerExponent[c, 2]; If[prevc>n&&c<n, h++]]; prevc=c]; h, {n, nterms}]

Prog

(Python)
def A349325(n):
    prevc = c = n
    h = 1
    while c > 1:
        if c % 2:
            c = (3*c+1) // 2
            if prevc < n and c > n: h += 1
        else:
            c //= 2
            if prevc > n and c < n: h += 1
        prevc = c
    return h
print([A349325(n) for n in range(1, 100)])
(PARI) f(n) = if (n%2, (3*n+1)/2, n/2); \\ A014682
a(n) = {my(nb=1, prec=n, next); while (prec != 1, next = f(prec); if ((next-n)*(prec-n) <0, nb++); prec = next; ); nb; } \\ Michel Marcus, Nov 16 2021

Crossrefs

Cf. A006370, A014682, A070168, A166245, A304030.

Sequence in context: A308897 A353293 A162190 * A134292 A344006 A353319

Adjacent sequences: A349322 A349323 A349324 * A349326 A349327 A349328

Keyword

nonn

Author

Paolo Xausa, Nov 15 2021

Status

approved