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 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). 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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. REFERENCES Jerash University Journal 2000-2001 LINKS Randall L. Rathbun, Discussi on of this sequence C. Rivera, Prime Puzzle 114 EXAMPLE With P=11 and c(0)=17 then {c(m)} is 17, 188, 94, 47, 518, 37, 408, 68, 34, 17,... CROSSREFS Prime complement of A058047. Cf. A057446, A057216, A057534, A057614, A058047. Sequence in context: A137238 A048521 A172071 * A241659 A038915 A166849 Adjacent sequences:  A058045 A058046 A058047 * A058049 A058050 A058051 KEYWORD nonn AUTHOR Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000 EXTENSIONS Edited by Henry Bottomley, Jun 14 2002 Corrected by T. D. Noe, Oct 25 2006 STATUS approved

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Last modified April 22 10:29 EDT 2019. Contains 322330 sequences. (Running on oeis4.)