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%I #16 May 25 2022 02:50:39
%S 2,11,13,17,19,23,31,37,43,47,53,59,61,67,71,73,83,97,101,103,113,131,
%T 137,139,151,163,167,173,181,193,197,223,227,229,233,239,241,251,257,
%U 263,269,271,277,281,283,293,313,331,347,353,367,373,379,383,389,401
%N 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).
%C 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.
%H Randall L. Rathbun, <a href="/A058047/a058047.txt">Discussion of this sequence</a>
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_114.htm">Puzzle 114. The Murad's generalization of the Collatz's sequences</a>, The Prime Puzzles and Problems Connection.
%e With P=11 and c(0)=17 then {c(m)} is 17, 188, 94, 47, 518, 37, 408, 68, 34, 17, ...
%Y Prime complement of A058047. Cf. A057446, A057216, A057534, A057614, A058047.
%K nonn
%O 1,1
%A Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000
%E Edited by _Henry Bottomley_, Jun 14 2002
%E Corrected by _T. D. Noe_, Oct 25 2006
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