

A349325


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


4



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
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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



FORMULA

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


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)
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 ((nextn)*(precn) <0, nb++); prec = next; ); nb; } \\ Michel Marcus, Nov 16 2021


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



