A359220 Number of steps to reach 0 from A359219(n) where A359219 are the starting numbers that require more iterations in the map x->A359194(x) than any smaller number.

0, 1, 2, 11, 12, 13, 19, 80, 81, 83, 7572, 7573, 7574, 7578, 7580, 664475, 664882, 3180929, 3180930, 3180931, 3181981, 3181988, 3182002, 3182226, 120796790, 556068798, 556068799, 556068871, 556068872, 572086553, 572086610, 1246707529, 1246707552, 1246707555, 1246707602

Offset: 1

Author

Joshua Searle, Dec 21 2022

Keywords

• nonn
• base

Comments

It is unknown whether all starting numbers reach 0; the next term, a(36) depends on whether 425720 ever reaches 0 (see A359207). It remains nonzero after 10^10 iterations.

A359207(425720) = 87037147316. Calculated by Tom Duff (12/16/22) - Joshua Searle, Jan 10 2023

a(4) - a(7) only differ by a small fraction of their starting terms. The same is true for the terms in the intervals a(8) - a(10), a(11) - a(15), a(16) - a(17) and a(18) - a(24). It may also be true for a(26) - a(29), a(30) - a(31) and a(32) - a(35).

Examples

			a(4) is the step count from the starting number A359219(4) = 3: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) -- 11 steps, hence a(4) = 11.
		

Links

• Joshua Searle, Collatz-inspired sequences

Crossrefs

Step counts of A359219.

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359219, A359221.

Programs

Python

from itertools import count, islice
def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def iters(n):
    i, fi = 0, n
    while fi != 0: i, fi = i+1, f(fi)
    return i
def agen(): # generator of terms
    record = -1
    for m in count(0):
        v = iters(m)
        if v > record: yield v; record = v
print(list(islice(agen(), 18))) # Michael S. Branicky, Dec 21 2022

Extensions

a(27) and beyond from Tom Duff (SeqFan mailing list, Dec 19 2022)