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A062282
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Number of permutations of n elements with an even number of fixed points.
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7
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1, 0, 2, 2, 16, 64, 416, 2848, 22912, 205952, 2060032, 22659328, 271913984, 3534877696, 49488295936, 742324422656, 11877190795264, 201912243453952, 3634420382302208, 69053987263479808, 1381079745270120448, 29002674650671480832, 638058842314774675456
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OFFSET
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0,3
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COMMENTS
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Let d(n) be the number of derangements of n elements (sequence A000166) then a(n) has the recursion: a(n) = d(n) + C(n,2)*d(n-2) + C(n,4)*d(n-4) + C(n,6)*d(n-6)... = A000166(n) + A000387(n) + A000475(n) + C(n,6)*d(n-6)... The E.g.f. for a(n) is: cosh(x) * exp(-x)/(1-x) and the asymptotic expression for a(n) is: a(n) ~ n! * (1 + 1/e^2)/2 i.e., as n goes to infinity the fraction of permutations that has an even number of fixed points is about (1 + 1/e^2)/2 = 0.567667...
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LINKS
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FORMULA
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a(n) = Sum_{k=0..[n/2]} Sum_{l=0..(n-2*k)} (-1)^l * n!/((2*k)! * l!).
More generally, e.g.f. for number of degree-n permutations with an even number of k-cycles is cosh(x^k/k)*exp(-x^k/k)/(1-x). - Vladeta Jovovic, Jan 31 2006
E.g.f.: 1/(1-x)/(x*E(0)+1), where E(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/E(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Dec 29 2013
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MATHEMATICA
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nn = 20; d = Exp[-x]/(1 - x); Range[0, nn]! CoefficientList[Series[Cosh[x] d, {x, 0, nn}], x] (* Geoffrey Critzer, Jan 14 2012 *)
Table[Sum[Sum[(-1)^j * n!/(j!*(2*k)!), {j, 0, n - 2*k}], {k, 0, Floor[n/2]}], {n, 0, 50}] (* G. C. Greubel, Aug 21 2017 *)
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PROG
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(PARI) for(n=0, 50, print1(sum(k=0, n\2, sum(j=0, n-2*k, (-1)^j*n!/(j!*(2*k)!))), ", ")) \\ G. C. Greubel, Aug 21 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 04 2001
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EXTENSIONS
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STATUS
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approved
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