A359208 Maximum value reached when starting from n during iteration of the map x->A359194(x) (binary complement of 3n), or -1 if infinite.

0, 1, 2, 300, 300, 5, 300, 10, 10, 300, 10, 300, 328536, 300, 21, 300, 300, 328536, 300, 300, 300, 21, 72, 328536, 300, 328536, 661, 328536, 123130640068522377168864228132316865867184046004226894, 40, 300, 328536, 328536

Offset: 0

Author

Joshua Searle, Dec 20 2022

Keywords

• nonn
• base

Comments

It is unknown whether any terms are -1. The next term a(33) is equal to a(28), a 54-digit number. a(425720) is >= 2.09 * 10^114778, unresolved after 10^10 iterations.

a(425720) = 7.14... * 10^179246. - Joshua Searle, Jan 10 2023

Examples

			a(3) = 300 because the largest term in the iterated sequence: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) is 300.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Joshua Searle, Collatz-inspired sequences

Crossrefs

Cf. A359194, A359207, A359209, A359215, A359218, A359219, A359220, A359221, A359222, A359255.

Programs

Mathematica

f[n_] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Table[Max@ NestWhileList[f, n, # != 0 &], {n, 0, 32}] (* Michael De Vlieger, Dec 21 2022 *)

PARI

f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
a(n) = my(x=n, m=n); while (m, m=f(3*m); if (m>x, x=m)); x; \\ Michel Marcus, Dec 21 2022

Python

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def a(n):
    i, fi, m = 0, n, n
    while fi != 0: i, fi, m = i+1, f(fi), max(m, fi)
    return m
print([a(n) for n in range(33)]) # Michael S. Branicky, Dec 20 2022