A058047 Generalized Collatz sequences: primes resulting in a cycle containing 1.

3, 5, 7, 29, 41, 79

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

Table of n, a(n) for n=0..5.

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

Cf. A057216, A057446, A057534, A057614, A058048, A106919.

Sequence in context: A076846 A046931 A154551 * A098860 A106920 A060273

Adjacent sequences: A058044 A058045 A058046 * A058048 A058049 A058050

Keyword

nonn,more

Author

Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000

Extensions

Edited by Frank Ellermann, Jan 20 2002

Status

approved