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A058048
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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).
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1
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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
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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With P=11 and c(0)=17 then {c(m)} is 17, 188, 94, 47, 518, 37, 408, 68, 34, 17, ...
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000
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EXTENSIONS
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STATUS
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approved
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