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A058047
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Generalized Collatz sequences: primes resulting in a cycle containing 1.
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2
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OFFSET
| 0,1
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COMMENTS
| For each prime P check the generalized Collatz sequence of each integer N > 1 defined by c(1) = N, c(n+1) = c(n) * P + 1 if F > P, else c(n+1) = c(n) / F, where F is the smallest factor of c(n), until c(n) = c(m) for n > m starts a cycle. If all c(i) > 1, then P does not belong to the sequence (and vice versa).
All terms are as yet only conjectures. Jeff Heleen checked the primes < 1000 and start points up to 10000000 (see Prime Puzzle 114 and example below). a(1)=3 is the ordinary Collatz problem. - Frank Ellermann, Jan 20 2002.
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REFERENCES
| Jerash University Journal, 2000-2001
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LINKS
| Randall L. Rathbun, Discussion of this sequence
C. Rivera, Puzzle 114
Eric Weisstein's World of Mathematics, Collatz problem
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EXAMPLE
| a(4) > 11, e.g.: 17, 17*11+1=188, 188/(2*2)=47, 47*11+1=518, 518/(2*7)=37, 37*11+1=408, 408/(2*2*2*3)=17 (cycle without 1).
For p = 29 e.g.: 17, 17*29+1=494, 494/(2*13*19)=1, 1*29+1=30, 30/30=1 (cycle with 1), no counter-example below 10000000.
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CROSSREFS
| Cf. A057446, A057216, A057534, A057614, A058048.
Sequence in context: A076846 A046931 A154551 * A098860 A106920 A060273
Adjacent sequences: A058044 A058045 A058046 * A058048 A058049 A058050
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KEYWORD
| nonn,more
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AUTHOR
| Murad A. AlDamen (Divisibility(AT)yahoo.com), Nov 17 2000
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EXTENSIONS
| Edited by hmdmhdfmhdjmzdtjmzdtzktdkztdjz(AT)gmail.com, Jan 20 2002.
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