

A162978


Number of fixed points in all alternating (i.e., downup) permutations of {1,2,...,n}.


3



1, 0, 1, 4, 15, 52, 257, 1272, 7679, 47864, 346113, 2604380, 22022143, 194053836, 1881735169, 18998097328, 207983607807, 2366490065968, 28880901505025, 365599818496116, 4922617151619071, 68612903386404260, 1010501269355233281, 15376572385777544744
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OFFSET

1,4


COMMENTS

a(n) = Sum_{k>=0} k*A162979(n,k).
a(2n+1) = A162977(2n+1).


LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..485
R. P. Stanley, Alternating permutations Talk slides.


FORMULA

a(2n) = E(2n) + (1)^n*E(0) + 2*Sum_{j=1..n1} (1)^j*E(2n2j), a(2n+1) = Sum_{j=0..n} (1)^j*E(2n+12j), where E(i) = A000111(i) are the Euler (or updown) numbers.


EXAMPLE

a(4)=4 because in the 5 (=A000111(4)) downup permutations of {1,2,3,4}, namely 4132, 3142, 2143, 4231, and 3241, we have a total of 1+0+0+2+1=4 fixed points.


MAPLE

E := sec(x)+tan(x): Eser := series(E, x = 0, 30): for n from 0 to 27 do E[n] := factorial(n)*coeff(Eser, x, n) end do: for n to 12 do a[2*n] := E[2*n]+(1)^n*E[0]+2*add((1)^j*E[2*n2*j], j = 1 .. n1) end do: for n from 0 to 12 do a[2*n+1] := add((1)^j*E[2*n+12*j], j = 0 .. n) end do: seq(a[n], n = 1 .. 25);


MATHEMATICA

a111[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1) (2^(n+1)  1) BernoulliB[n+1])/(n+1)]];
a[n_?EvenQ] := With[{m = n/2}, a111[2m] + (1)^m a111[0] + 2Sum[(1)^j a111[2m  2j], {j, 1, m1}]];
a[n_?OddQ] := With[{m = (n1)/2}, Sum[(1)^j a111[2m+12j], {j, 0, m}]];
Array[a, 25] (* JeanFrançois Alcover, Jul 24 2018 *)


CROSSREFS

Cf. A000111, A162977, A162979.
Sequence in context: A208722 A057332 A230623 * A171309 A210781 A303271
Adjacent sequences: A162975 A162976 A162977 * A162979 A162980 A162981


KEYWORD

nonn


AUTHOR

Emeric Deutsch, Aug 06 2009


STATUS

approved



