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

Author

Paolo Xausa, Nov 15 2021

Keywords

• nonn

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.

Examples

			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.
		

References

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

Links

• 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

Crossrefs

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

Programs

Mathematica

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

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

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)])

Formula

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

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