OFFSET
0,5
COMMENTS
Row n has 1+floor(n/2) entries. T(2n-1,0) = T(2n,0) = T(2n,n) = (n!)^2 = A001044(n).
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
LINKS
S. Kitaev and J. Remmel, Classifying descents according to parity, Annals of Combinatorics, 11, 2007, 173-193.
FORMULA
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).
EXAMPLE
T(4,2) = 4 because we have 2143, 4213, 3421 and 4321.
Triangle starts:
1;
1;
1, 1;
4, 2;
4, 16, 4;
36, 72, 12;
36, 324, 324, 36;
...
MAPLE
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
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Nov 22 2007
EXTENSIONS
T(0,0)=1 prepended by Alois P. Heinz, Jul 12 2024
STATUS
approved