%I #20 Jul 12 2024 22:00:39
%S 1,1,1,1,4,2,4,16,4,36,72,12,36,324,324,36,576,2592,1728,144,576,9216,
%T 20736,9216,576,14400,115200,172800,57600,2880,14400,360000,1440000,
%U 1440000,360000,14400,518400,6480000,17280000,12960000,2592000,86400
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k even entries that are followed by a smaller entry (n>=0, k>=0).
%C Row n has 1+floor(n/2) entries. T(2n-1,0) = T(2n,0) = T(2n,n) = (n!)^2 = A001044(n).
%C This descent statistic is equidistributed on the symmetric group S_n with a multiplicative 2-excedance statistic - see A136715 for details. - _Peter Bala_, Jan 23 2008
%H S. Kitaev and J. Remmel, <a href="http://dx.doi.org/10.1007/s00026-007-0313-2">Classifying descents according to parity</a>, Annals of Combinatorics, 11, 2007, 173-193.
%F T(2n,k) = [n!*C(n,k)]^2; T(2n+1,k) = [(n+1)!*C(n,k)]^2/(k+1). See the Kitaev & Remmel reference for recurrence relations (Sec. 3).
%e T(4,2) = 4 because we have 2143, 4213, 3421 and 4321.
%e Triangle starts:
%e 1;
%e 1;
%e 1, 1;
%e 4, 2;
%e 4, 16, 4;
%e 36, 72, 12;
%e 36, 324, 324, 36;
%e ...
%p R[0]:=1:R[1]:=1: R[2]:=1+t: for n to 5 do R[2*n+1]:=sort(expand((1-t)* (diff(R[2*n], t))+(2*n+1)*R[2*n])): R[2*n+2]:=sort(expand(t*(1-t)*(diff(R[2*n+1], t))+(1+(2*n+1)*t)*R[2*n+1])) end do: for n from 0 to 11 do seq(coeff(R[n], t, j), j=0..floor((1/2)*n)); end do; # yields sequence in triangular form
%t T[n_,k_]:=If[EvenQ[n],Floor[(n/2)!Binomial[n/2,k]]^2, Floor[((n+1)/2)!Binomial[(n-1)/2,k]]^2/(k+1)]; Table[T[n,k],{n,11},{k,0,Floor[n/2]}]//Flatten (* _Stefano Spezia_, Jul 12 2024 *)
%Y Bisection of column k=0 gives A001044 (even part).
%Y Row sums give A000142.
%Y Cf. A134435, A136715.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Nov 22 2007
%E T(0,0)=1 prepended by _Alois P. Heinz_, Jul 12 2024