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A060171
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Number of orbits of length n under a map whose periodic points seem to be counted by A006953.
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11
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12, 54, 80, 30, 24, 5400, 0, 990, 1568, 636, 24, 2720, 0, 240, 5704, 510, 0, 3835776, 0, 26724, 3600, 108, 24, 89760, 0, 240, 1064, 120, 24, 113569300, 0, 510, 11752, 0, 264, 278281640
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OFFSET
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1,1
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COMMENTS
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The sequence A006953 seems to record the number of points of period n under a map. The number of orbits of length n for this map gives the sequence above.
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LINKS
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FORMULA
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a(n) = (1/n)* Sum_{d|n} mu(d)*A006953(n/d).
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EXAMPLE
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u(3) = 80 since a map whose periodic points are counted by A006953 has 12 fixed points and 252 points of period 3, hence 80 orbits of length 3.
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PROG
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(PARI) a(n) = sumdiv(n, d, moebius(d)*denominator(bernfrac(2*n/d)/(2*n/d)))/n; \\ Michel Marcus, Sep 10 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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