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A380993 Irregular triangular array read by rows. T(n,k) is the number of ternary words of length n containing at least one copy of each letter and having exactly k inversions, n>=3, 0<=k<=floor(n^2/3).

%I A380993 #20 Feb 14 2025 11:14:05
%S A380993 1,2,2,1,3,6,9,9,6,3,6,12,21,27,30,24,18,9,3,10,20,38,55,74,81,80,69,
%T A380993 53,34,17,8,1,15,30,60,93,138,174,210,216,219,195,165,120,84,48,27,9,
%U A380993 3,21,42,87,141,222,303,405,480,546,579,588,552,498,414,324,240,162,99,54,27,9,3
%N A380993 Irregular triangular array read by rows.  T(n,k) is the number of ternary words of length n containing at least one copy of each letter and having exactly k inversions, n>=3, 0<=k<=floor(n^2/3).
%H A380993 Alois P. Heinz, <a href="/A380993/b380993.txt">Rows n = 3..50, flattened</a>
%F A380993 Sum_{n>=0} Sum_{k>=0} T(n,k)*q^k*x^n/B(n) = (e(x)-1)^3 where B(n) = Product_{i=1..n} (q^i-1)/(q-1) and e(x) = Sum_{n>=0} x^n/B(n).
%F A380993 Sum_{k=0..floor(n^2/3)} (-1)^k * T(n,k) = A056454(n). - _Alois P. Heinz_, Feb 12 2025
%e A380993 Triangle T(n,k) begins:
%e A380993    1,  2,  2,  1;
%e A380993    3,  6,  9,  9,  6,  3;
%e A380993    6, 12, 21, 27, 30, 24, 18,  9,  3;
%e A380993   10, 20, 38, 55, 74, 81, 80, 69, 53, 34, 17, 8, 1;
%e A380993   ...
%e A380993 T(4,2) = 9 because we have: {0, 1, 2, 0}, {0, 2, 0, 1}, {0, 2, 1, 1}, {0, 2, 2, 1}, {1, 0, 0, 2}, {1, 0, 2, 1}, {1, 1, 0, 2}, {1, 2, 0, 2}, {2, 0, 1, 2}.
%p A380993 b:= proc(n, l) option remember; `if`(n=0, `if`(nops(subs(0=
%p A380993       [][], l))=3, 1, 0), add(expand(x^([0, l[1], l[1]+l[2]][j])*
%p A380993       b(n-1, subsop(j=`if`(j=3, 1, l[j]+1), l))), j=1..3))
%p A380993     end:
%p A380993 T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, [0$3])):
%p A380993 seq(T(n), n=3..10);  # _Alois P. Heinz_, Feb 12 2025
%t A380993 nn = 8; B[n_] := FunctionExpand[QFactorial[n, u]];
%t A380993 e[z_] := Sum[z^n/B[n], {n, 0, nn}];
%t A380993 Drop[Map[CoefficientList[#, u] &,
%t A380993    Map[Normal[Series[#, {u, 0, Binomial[nn, 2]}]] &,
%t A380993     Table[B[n], {n, 0, nn}] CoefficientList[
%t A380993       Series[(e[z] - 1)^3, {z, 0, nn}], z]]], 3] // Grid
%Y A380993 Cf. A056454, A129529, A001117 (row sums).
%K A380993 nonn,tabf
%O A380993 3,2
%A A380993 _Geoffrey Critzer_, Feb 11 2025