

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
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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 20002001


LINKS

Table of n, a(n) for n=1..56.
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



