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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)).

%I A161121 #15 Feb 01 2023 07:07:13
%S A161121 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,
%T A161121 5040,0,52920,0,66150,0,11025,40320,0,564480,0,1058400,0,352800,0,
%U A161121 11025,362880,0,6531840,0,17146080,0,9525600,0,893025,3628800,0,81648000,0
%N 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)).
%C A161121 Row n contains 1 + 2*floor(n/2) terms.
%C A161121 Sum of row n = (2n-1)!! (A001147).
%C A161121 a(n,0) = n! (A000142).
%C A161121 a(2n,2n) = A001818(n).
%C A161121 Sum_{k>=0} k*T(n,k) = n*(n-1)*(2n-3)!! = A161122(n).
%H A161121 Robert Israel, <a href="/A161121/b161121.txt">Table of n, a(n) for n = 0..10081</a> (rows 0 to 141, flattened)
%F A161121 T(n,k) = (n-k)!*binomial(n,k)^2*((k-1)!!)^2 if k is even; T(n,k) = 0 if k is odd.
%e A161121 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).
%e A161121 Triangle starts:
%e A161121     1;
%e A161121     1;
%e A161121     2,   0,   1;
%e A161121     6,   0,   9;
%e A161121    24,   0,  72,   0,   9;
%e A161121   120,   0, 600,   0, 225;
%p A161121 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
%t A161121 T[n_, k_] := If[EvenQ[k], (n-k)! Binomial[n, k]^2 ((k-1)!!)^2, 0];
%t A161121 Table[T[n, k], {n, 0, 10}, {k, 0, 2 Quotient[n, 2]}] // Flatten (* _Jean-François Alcover_, Feb 01 2023 *)
%Y A161121 Cf. A000142, A001147, A001818, A161119, A161120, A161122.
%K A161121 nonn,tabf
%O A161121 0,3
%A A161121 _Emeric Deutsch_, Jun 02 2009