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A152661 Number of permutations of [n] for which the first two entries have the same parity (n>=2). 1
0, 2, 8, 48, 288, 2160, 17280, 161280, 1612800, 18144000, 217728000, 2874009600, 40236134400, 610248038400, 9763968614400, 167382319104000, 3012881743872000, 57621363351552000, 1152427267031040000, 24329020081766400000, 535238441798860800000 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,2

COMMENTS

a(n) is also the number of 3-term arithmetic progressions of consecutive entries in all permutations of {1,2,...,n}. Example: a(4)=8 because we have 12'3'4, 412'3, 143'2, 23'41, 32'14, 43'2'1 (the mid-terms of the arithmetic progressions are marked). [Emeric Deutsch, Aug 31 2009]

LINKS

Table of n, a(n) for n=2..22.

S. Tanimoto, Combinatorial study on the group of parity alternating permutations, arXiv:0812.1839 [math.CO], 2008-2017.

FORMULA

a(n) = A152660(n,1).

a(2n) = 2*(n!)^2*binomial(2*n-2,n);

a(2n+1) = n!*(n+1)!*binomial(2n,n-1).

Conjecture: (-n+3)*a(n) +2*(n-2)*a(n-1) +(n-1)*(n-2)*(n-3)*a(n-2)=0. - R. J. Mathar, Apr 20 2015

Conjecture: a(n) = 2*A077613(n-1). - R. J. Mathar, Apr 20 2015

EXAMPLE

a(4)=8 because we have 1324, 1342, 3124, 3142, 2413, 2431, 4213 and 4231.

MAPLE

a := proc (n) if `mod`(n, 2) = 0 then 2*factorial((1/2)*n)^2*binomial(n-2, (1/2)*n) else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial(n-1, (1/2)*n-3/2) end if end proc: seq(a(n), n = 2 .. 22);

MATHEMATICA

a[n0_?EvenQ] := With[{n = n0/2}, 2 (n!)^2*Binomial[2*n - 2, n]];

a[n1_?OddQ] := With[{n = (n1 - 1)/2}, n! (n + 1)! Binomial[2 n, n - 1]];

Table[a[n], {n, 2, 22}] (* Jean-Fran├žois Alcover, Nov 28 2017 *)

CROSSREFS

Cf. A152660.

Sequence in context: A009693 A192251 A104190 * A177066 A228568 A007170

Adjacent sequences:  A152658 A152659 A152660 * A152662 A152663 A152664

KEYWORD

nonn

AUTHOR

Emeric Deutsch, Dec 12 2008

STATUS

approved

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Last modified December 7 17:41 EST 2019. Contains 329847 sequences. (Running on oeis4.)