%I #23 Apr 04 2024 07:53:42
%S 1,0,1,1,1,2,2,2,12,8,4,48,36,24,12,360,216,108,36,2160,1440,864,432,
%T 144,20160,11520,5760,2304,576,161280,100800,57600,28800,11520,2880,
%U 1814400,1008000,504000,216000,72000,14400,18144000,10886400,6048000,3024000
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} for which k is the maximal number of initial odd entries (0 <= k <= ceiling(n/2)).
%C Sum of entries in row n is n! (A000142).
%C Row n has 1 + ceiling(n/2) entries.
%C T(n,0) = A052591(n-1) for n>=1.
%C Sum_{k=0..ceiling(n/2)} k*T(n,k) = A152663(n).
%H Alois P. Heinz, <a href="/A152662/b152662.txt">Rows n = 0..200, flattened</a>
%F T(2n+1,k) = n*k!*(2n-k)!*binomial(n+1,k) (n>= 1);
%F T(2n,k) = n*k!*(2n-1-k)!*binomial(n,k).
%F From _Alois P. Heinz_, Apr 02 2024: (Start)
%F Sum_{k>=0} (k+1) * T(n,k) = A256881(n+1).
%F T(n,ceiling(n/2)) = A010551(n). (End)
%e T(3,0)=2 because we have 213 and 231.
%e T(4,2)=4 because we have 1324, 1342, 3124 and 3142.
%e Triangle starts:
%e 1;
%e 0, 1;
%e 1, 1;
%e 2, 2, 2;
%e 12, 8, 4;
%e 48, 36, 24, 12;
%e 360, 216, 108, 36;
%e ...
%p T := proc (n, k) if n=0 and k=0 then 1 elif n = 1 and k = 0 then 0 elif n = 1 and k = 1 then 1 elif `mod`(n, 2) = 1 then (1/2)*(n-1)*binomial((1/2)*n+1/2, k)*factorial(k)*factorial(n-1-k) else (1/2)*n*binomial((1/2)*n, k)*factorial(k)*factorial(n-1-k) end if end proc: for n from 0 to 11 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
%t T[n_, k_] := T[n, k] = Which[n == 0 && k == 0, 1, n == 1 && k == 1, 1, OddQ[n], (n - 1)/2*k!*(n - k - 1)!*Binomial[(n - 1)/2 + 1, k], True, n/2*k!*(n - k - 1)!*Binomial[n/2, k]];
%t Table[T[n, k], {n, 0, 11}, {k, 0, Ceiling[n/2]}] // Flatten (* _Jean-François Alcover_, Apr 04 2024 *)
%Y Cf. A000142, A010551, A052591, A152663, A152664, A256881.
%K nonn,tabf
%O 0,6
%A _Emeric Deutsch_, Dec 13 2008
%E T(0,0)=1 prepended by _Alois P. Heinz_, Apr 02 2024