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A060165
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Number of orbits of length n under the map whose periodic points are counted by A000984.
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18
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2, 2, 6, 16, 50, 150, 490, 1600, 5400, 18450, 64130, 225264, 800046, 2865226, 10341150, 37566720, 137270954, 504171432, 1860277042, 6892317200, 25631327190, 95640829922, 357975249026, 1343650040256, 5056424257500, 19073789328750, 72108867614796
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OFFSET
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1,1
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COMMENTS
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The sequence A000984 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.
The number of n-cycles in the graph of overlapping m-permutations where n <= m. - Richard Ehrenborg, Dec 10 2013
Apparently the number of Lyndon words of length n with a 4-letter alphabet (see A027377) where the first letter of the alphabet appears with the same frequency as the second of the alphabet. E.g a(1)=2 counts the words (2), (3), a(2)= 2 counts (01) (23), a(3)=6 counts (021) (031) (012) (013) (223) (233). R. J. Mathar, Nov 04 2021
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LINKS
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FORMULA
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G.f.: 2 * Sum_{k>=1} mu(k)*log((1 - sqrt(1 - 4*x^k))/(2*x^k))/k. - Ilya Gutkovskiy, May 18 2019
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EXAMPLE
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a(5) = 50 because if a map has A000984 as its periodic points, then it would have 2 fixed points and 252 points of period 5, hence 50 orbits of length 5.
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MAPLE
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with(numtheory):
a:= n-> add(mobius(n/d)*binomial(2*d, d), d=divisors(n))/n:
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MATHEMATICA
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a[n_] := (1/n)*Sum[MoebiusMu[d]*Binomial[2*n/d, n/d], {d, Divisors[n]}]; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, Jul 16 2015 *)
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PROG
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(Python)
from sympy import mobius, binomial, divisors
def a(n): return sum(mobius(n//d) * binomial(2*d, d) for d in divisors(n))//n
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CROSSREFS
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Cf. A000984, A007727, A060164, A060166, A060167, A060168, A060169, A060170, A060171, A060172, A060173.
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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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