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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
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.
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.
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
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