A161121 - OEIS

A161121

Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k cycles with entries of the same parity (0 <= k <= 2*floor(n/2)).

1, 1, 2, 0, 1, 6, 0, 9, 24, 0, 72, 0, 9, 120, 0, 600, 0, 225, 720, 0, 5400, 0, 4050, 0, 225, 5040, 0, 52920, 0, 66150, 0, 11025, 40320, 0, 564480, 0, 1058400, 0, 352800, 0, 11025, 362880, 0, 6531840, 0, 17146080, 0, 9525600, 0, 893025, 3628800, 0, 81648000, 0

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OFFSET

0,3

COMMENTS

Row n contains 1 + 2*floor(n/2) terms.

Sum of row n = (2n-1)!! (A001147).

a(n,0) = n! (A000142).

a(2n,2n) = A001818(n).

Sum_{k>=0} k*T(n,k) = n*(n-1)*(2n-3)!! = A161122(n).

LINKS

Robert Israel, Table of n, a(n) for n = 0..10081 (rows 0 to 141, flattened)

FORMULA

T(n,k) = (n-k)!*binomial(n,k)^2*((k-1)!!)^2 if k is even; T(n,k) = 0 if k is odd.

EXAMPLE

T(3,2)=9 because we have (12)(35)(46), (14)(26)(35), (16)(24)(35), (23)(15)(46), (25)(13)(46), (34)(15)(26), (36)(15)(24), (45)(13)(26), (56)(13)(24).

Triangle starts:

2, 0, 1;

6, 0, 9;

24, 0, 72, 0, 9;

120, 0, 600, 0, 225;

MAPLE

T := proc (n, k) if n < k then 0 elif `mod`(k, 2) = 0 then binomial(n, k)^2*factorial(n-k)*(product(2*j-1, j = 1 .. (1/2)*k))^2 else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. 2*floor((1/2)*n)) end do; # yields sequence in triangular form

MATHEMATICA

T[n_, k_] := If[EvenQ[k], (n-k)! Binomial[n, k]^2 ((k-1)!!)^2, 0];

Table[T[n, k], {n, 0, 10}, {k, 0, 2 Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Feb 01 2023 *)

CROSSREFS

Cf. A000142, A001147, A001818, A161119, A161120, A161122.

Sequence in context: A321686 A055925 A276436 * A021500 A175197 A367795

Adjacent sequences: A161118 A161119 A161120 * A161122 A161123 A161124

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, Jun 02 2009

STATUS

approved