A058048 For each prime P consider the generalized Collatz sequence of each integer N > 1 defined by c(0) = N, c(m+1) = c(m) * P + 1 if F > P, else c(m+1) = c(m) / F, where F is the smallest factor of c(m), until the sequence cycles. If all c(i) > 1 for some starting number N then P belongs to the sequence (and vice versa).

2, 11, 13, 17, 19, 23, 31, 37, 43, 47, 53, 59, 61, 67, 71, 73, 83, 97, 101, 103, 113, 131, 137, 139, 151, 163, 167, 173, 181, 193, 197, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 313, 331, 347, 353, 367, 373, 379, 383, 389, 401

Offset: 1

Author

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

Keywords

• nonn

Comments

Missing primes are as yet only conjectures. Jeff Heleen checked the primes < 1000 and start points up to 10000000 (see Prime Puzzle 114 and example below). P=3 is the ordinary Collatz problem.

Examples

			With P=11 and c(0)=17 then {c(m)} is 17, 188, 94, 47, 518, 37, 408, 68, 34, 17, ...
		

Links

• Randall L. Rathbun, Discussion of this sequence
• Carlos Rivera, Puzzle 114. The Murad's generalization of the Collatz's sequences, The Prime Puzzles and Problems Connection.

Crossrefs

Prime complement of A058047. Cf. A057446, A057216, A057534, A057614, A058047.

Extensions

Edited by Henry Bottomley, Jun 14 2002

Corrected by T. D. Noe, Oct 25 2006