OFFSET
0,1
COMMENTS
For each prime P check the generalized Collatz sequence of each integer N > 1 defined by c(1) = N, c(n+1) = c(n) * P + 1 if F > P, otherwise c(n+1) = c(n) / F, where F is the smallest factor of c(n), until c(n) = c(m) for n > m starts a cycle. If all c(i) > 1, then P does not belong to the sequence (and vice versa).
All terms are as yet only conjectures. Jeff Heleen checked the primes < 1000 and start points up to 10000000 (see Prime Puzzle 114 and example below). a(1)=3 is the ordinary Collatz problem. - Frank Ellermann, Jan 20 2002
The jOEIS program uses start points up to 10^8 and yields [3, 5, 7, 19, 29, 41, 43*, 53, 71*, 79, 89*, 103*, 107, 109*, 127, 131*, 137] followed by [139, 149, 157, 179, 191, 197, 199, 211, 227, ...]. The terms in the first list without asterisks agree with A106919. - Georg Fischer, Jun 17 2023
LINKS
Sean A. Irvine, Java program in the jOEIS project.
Randall L. Rathbun, Discussion of this sequence
Carlos Rivera, Puzzle 114. The Murad's generalization of the Collatz's sequences, The Prime Puzzles & Problems Connection.
Eric Weisstein's World of Mathematics, Collatz problem
EXAMPLE
a(4) > 11, e.g.: 17, 17*11 + 1 = 188, 188/(2*2) = 47, 47*11 + 1 = 518, 518/(2*7) = 37, 37*11 + 1 = 408, 408/(2*2*2*3) = 17 (cycle without 1).
For p = 29 e.g.: 17, 17*29 + 1 = 494, 494/(2*13*19) = 1, 1*29 + 1 = 30, 30/30 = 1 (cycle with 1), no counterexample below 10000000.
PROG
(Java) Cf. link to the program in the jOEIS project.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000
EXTENSIONS
Edited by Frank Ellermann, Jan 20 2002
STATUS
approved