A088312 revision #28
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A088312
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Number of sets of lists (cf. A000262) with even number of lists.
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4
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1, 0, 1, 6, 37, 260, 2101, 19362, 201097, 2326536, 29668681, 413257790, 6238931821, 101415565836, 1765092183037, 32734873484250, 644215775792401, 13404753632014352, 293976795292186897, 6775966692145553526, 163735077313046119861, 4138498600079573989140
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OFFSET
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0,4
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COMMENTS
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a(2*n) is odd ; a(2*n+1) is even.
If k is odd then k*(k-1) divides a(k). Consequently, 6 divides a(6*n+3), 10 divides a(10*n+5), 14 divides a(14*n+7), and in general, if k is odd then 2*k divides a(2*k*n + k).
For a positive integer k, a(n+2*k) - a(n) is divisible by k. Thus the sequence obtained by taking a(n) modulo k is purely periodic with period 2*k. Calculation suggests that when k is even the exact period equals k, and when k is odd the exact period equals 2*k. (End)
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LINKS
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FORMULA
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E.g.f.: cosh(x/(1-x)).
a(n) = Sum_{k=1..floor(n/2)} n!/(2*k)!*binomial(n-1,2*k-1).
a(n) ~ 2^(-3/2) * n^(n-1/4) * exp(2*sqrt(n)-n-1/2). - Vaclav Kotesovec, Jul 04 2015
a(n+4) - 2*(2*n+5)*a(n+3) + (6*n^2+24*n+23)*a(n+2) - 2*(n+1)*(n+2)*(2*n+3)*a(n+1) + n*(n+1)^2*(n+2)*a(n) = 0. - Emanuele Munarini, Sep 03 2017
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MAPLE
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b:= proc(n, t) option remember; `if`(n=0, t, add(
b(n-j, 1-t)*binomial(n-1, j-1)*j!, j=1..n))
end:
a:= n-> b(n, 1):
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MATHEMATICA
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Rest[Rest[CoefficientList[Series[Cosh[x/(1-x)], {x, 0, 20}], x] * Range[0, 20]!]] (* Vaclav Kotesovec, Jul 04 2015 *)
Table[Sum[n!/(2*k)! Binomial[n - 1, 2*k - 1], {k, 0, Floor[n/2]}], {n, 0, 12}] (* Emanuele Munarini, Sep 03 2017 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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proposed
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