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

1,3

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).

Links

Table of n, a(n) for n=1..35.

Joshua Searle, Collatz-inspired sequences

Example

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.

Prog

(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

Crossrefs

Step counts of A359219.

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

Sequence in context: A108199 A137904 A175414 * A059192 A087339 A345922

Adjacent sequences: A359217 A359218 A359219 * A359221 A359222 A359223

Keyword

nonn,base

Author

Joshua Searle, Dec 21 2022

Extensions

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

Status

approved