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A161123 Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k inversions (0 <= k <= n(2n-1)).

%I A161123 #11 Aug 23 2024 10:48:56
%S A161123 1,0,1,0,0,1,0,1,0,1,0,0,0,1,0,2,0,3,0,3,0,3,0,2,0,1,0,0,0,0,1,0,3,0,
%T A161123 6,0,9,0,12,0,14,0,15,0,14,0,12,0,9,0,6,0,3,0,1,0,0,0,0,0,1,0,4,0,10,
%U A161123 0,19,0,31,0,45,0,60,0,74,0,86,0,94,0,97,0,94,0,86,0,74,0,60,0,45,0,31,0,19
%N A161123 Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k inversions (0 <= k <= n(2n-1)).
%C A161123 Sum of entries in row n is (2n-1)!! = A001147(n).
%C A161123 Row n has 1 + 2n(n-1) entries.
%C A161123 Sum_{k>=0} k*T(n,k) = (2n-1)!!*n^2 = A161124(n).
%C A161123 A128080 is the same triangle with the 0's deleted.
%F A161123 Generating polynomial of row n is P_n(q) = (q/(1-q^2))^n*Product_{j=1..n}(1-q^(4j-2)).
%e A161123 T(3,11)=3 because we have 465132, 546213, and 632541.
%e A161123 Triangle starts:
%e A161123   1;
%e A161123   0, 1;
%e A161123   0, 0, 1, 0, 1, 0, 1;
%e A161123   0, 0, 0, 1, 0, 2, 0, 3, 0, 3, 0, 3, 0, 2, 0, 1;
%p A161123 f := proc (n) options operator, arrow: q^n*(product(1-q^(4*j-2), j = 1 .. n))/(1-q^2)^n end proc: for n from 0 to 4 do P[n] := sort(expand(simplify(f(n)))) end do: for n from 0 to 4 do seq(coeff(P[n], q, j), j = 0 .. n*(2*n-1)) end do; # yields sequence in triangular form
%t A161123 P[n_] := P[n] = q^n*Product[1 - q^(4j - 2), {j, 1, n}]/(1 - q^2)^n // Expand // Simplify;
%t A161123 T[n_, k_] := Coefficient[P[n], q, k];
%t A161123 Table[T[n, k], {n, 0, 5}, {k, 0, n (2n - 1)}] // Flatten (* _Jean-François Alcover_, Aug 23 2024 *)
%Y A161123 Cf. A001147, A128080, A161124.
%K A161123 nonn,tabf
%O A161123 0,16
%A A161123 _Emeric Deutsch_, Jun 05 2009