A359219 Starting numbers that require more iterations of the map x->A359194(x) (binary complement of 3n) to reach 0 than any smaller number.

0, 1, 2, 3, 4, 9, 11, 12, 17, 23, 28, 33, 74, 86, 180, 227, 350, 821, 3822, 4187, 5561, 6380, 6398, 22174, 22246, 26494, 34859, 49827, 70772, 103721, 104282, 204953, 213884, 225095, 407354, 425720

Offset: 1

Author

Joshua Searle, Dec 21 2022

Keywords

• nonn
• base
• more

Comments

425720 after 10^10 iterations has not yet reached 0 and in general it is unknown whether every starting number does reach 0.

A359207(425720) = 87037147316. - Martin Ehrenstein, Jan 02 2023

Examples

			3 is a term because it requires 11 iterations to reach 0, which is more than any starting number less than 3.
0: (0) -- 0 terms
1: (1, 0) -- 1 term
2: (2, 1, 0) -- 2 terms
3: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) -- 11 terms.
		

Links

• Joshua Searle, Collatz-inspired sequences

Crossrefs

Cf. A035327, A359194, A359207, A359208, A359209, A359215, A359218, A359220, A359221, A359222.

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 m; record = v
print(list(islice(agen(), 18))) # Michael S. Branicky, Dec 21 2022

Extensions

a(27)-a(36) from Tom Duff (SeqFan mailing list, Dec 19 2022)