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A056957
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In repeated iterations of function m->m/2 if m even, m->3m-1 if m odd, a(n) is minimum value achieved if starting from n.
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1
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1, 1, 1, 1, 5, 1, 5, 1, 5, 5, 1, 1, 5, 5, 1, 1, 17, 5, 5, 5, 17, 1, 17, 1, 17, 5, 5, 5, 1, 1, 17, 1, 17, 17, 5, 5, 17, 5, 1, 5, 17, 17, 1, 1, 17, 17, 5, 1, 17, 17, 5, 5, 1, 5, 17, 5, 1, 1, 1, 1, 17, 17, 5, 1, 1, 17, 17, 17, 1, 5, 1, 5, 17, 17, 5, 5, 1, 1, 1, 5, 5, 17, 17, 17, 1, 1, 1, 1, 5, 17
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OFFSET
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1,5
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COMMENTS
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At least for n<10000, the only possible cycles reached include 1,2,1,..., 5,14,7,20,10,5,... and 17,50,25,74,37,110,55,164,82,41,122,61,182,91,272,136,68,34,17,... For n<5 only the first occurs, while for n<17 only the first two occur.
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LINKS
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FORMULA
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a(2n) = a(n)
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EXAMPLE
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a(9)=5 since iteration starts: 9, 26, 13, 38, 19, 56, 28, 14, 7, 20, 10, 5, 14, 7, 20, 10, 5, ... and 5 is the smallest value
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CROSSREFS
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Cf. A001281. If n is in A039500 then a(n)=1, if n is in A039501 then a(n)=5, if n is in A039502 then a(n)=17. If n is negative then this becomes the 3x+1 problem and the minimum values become those which are most negative (i.e. maximum absolute values) as in A056959.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Edited by Bryce Herdt (mathidentity(AT)yahoo.com), Apr 18 2010
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STATUS
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approved
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