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%I #23 Jul 12 2024 16:16:02
%S 1,1,2,2,4,12,12,12,72,36,144,432,144,144,1728,2592,576,2880,17280,
%T 17280,2880,2880,57600,172800,115200,14400,86400,864000,1728000,
%U 864000,86400,86400,2592000,12960000,17280000,6480000,518400
%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k odd entries that are followed by a smaller entry (n >= 0, k >= 0).
%C Row n has ceiling(n/2) entries (for n>0). T(2n,0) = T(2n+1,0) = n!*(n+1)! = A010790(n).
%C T(n,k) is also the number of permutations of {1,2,...,n} having k adjacent pairs of the form (odd, odd) (0 <= k <= ceiling(n,2)-1). Example: T(3,1)=4 because we have 132, 213, 312 and 231. - _Emeric Deutsch_, Dec 14 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!)^2*C(n-1,k) C(n+1,k+1); T(2n+1,k) = n!(n+1)! * C(n,k) * C(n+1,k).
%e T(3,1) = 4 because we have 132, 312, 231 and 321.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 2, 4;
%e 12, 12;
%e 12, 72, 36;
%e 144, 432, 144;
%e ...
%p T:=proc(n, k) if `mod`(n, 2)=0 then binomial((1/2)*n-1, k)*binomial((1/2)* n+1, k+1)*factorial((1/2)*n)^2 elif `mod`(n, 2)=1 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, k)*binomial((1/2)* n+1/2, k) else 0 end if end proc: for n from 0 to 11 do seq(T(n, k), k=0..max(0,ceil((1/2)*n)-1)) end do; # yields sequence in triangular form
%t T[n_,k_]:=If[EvenQ[n],((n/2)!)^2Binomial[n/2-1,k]Binomial[n/2+1,k+1], ((n-1)/2)!((n+1)/2)!Binomial[(n-1)/2,k]Binomial[(n+1)/2,k]]; Table[T[n,k],{n,11},{k,0,Floor[(n-1)/2]}]//Flatten (* _Stefano Spezia_, Jul 12 2024 *)
%Y Bisection of column k=0 gives A010790.
%Y Row sums give A000142.
%Y Cf. A134434.
%K nonn,tabf,changed
%O 0,3
%A _Emeric Deutsch_, Nov 22 2007
%E T(0,0)=1 prepended by _Alois P. Heinz_, Jul 12 2024
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