%I #28 Dec 20 2021 05:53:01
%S 1,0,1,1,0,2,0,9,0,6,9,0,72,0,24,0,225,0,600,0,120,225,0,4050,0,5400,
%T 0,720,0,11025,0,66150,0,52920,0,5040,11025,0,352800,0,1058400,0,
%U 564480,0,40320,0,893025,0,9525600,0,17146080,0,6531840,0,362880,893025,0,44651250,0,238140000,0,285768000,0,81648000,0,3628800
%N 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 opposite parities (0 <= k <= n).
%C T(n,k) is the number of basis elements in the order-n Brauer algebra that have propagation number k. - _John M. Campbell_, Dec 08 2021
%H Andrew Howroyd, <a href="/A161119/b161119.txt">Table of n, a(n) for n = 0..1325</a> (rows 0..50)
%F T(n,k) = k!*binomial(n,k)^2*(n-k-1)!!^2 if n-k is even; T(n,k) = 0 if n-k is odd.
%F Sum of row n = (2n-1)!! = A001147(n).
%F T(n,n) = n! = A000142(n).
%F T(2n,0) = A001818(n).
%F Sum_{k>=0} k*T(n,k) = n^2*(2n-3)!! = A161120(n).
%e T(3,1)=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 Triangle starts:
%e 1;
%e 0, 1;
%e 1, 0, 2;
%e 0, 9, 0, 6;
%e 9, 0, 72, 0, 24;
%p T := proc (n, k) if `mod`(n-k, 2) = 1 then 0 else binomial(n, k)^2*factorial(k)*(product(2*j-1, j = 1 .. (1/2)*n-(1/2)*k))^2 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
%o (PARI) dfo(n) = if (n<0, (-1)^n/dfo(-n), (2*n)! / n! / 2^n); \\ A001147
%o T(n,k) = if ((n-k)%2, 0, k!*binomial(n,k)^2*dfo((n-k)/2)^2);
%o row(n) = vector(n+1, k, T(n,k-1)) \\ _Michel Marcus_, Dec 09 2021
%Y Cf. A000142, A001147, A001818, A161120, A161121, A161122.
%K nonn,tabl
%O 0,6
%A _Emeric Deutsch_, Jun 02 2009