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%I #10 Mar 11 2020 23:52:19
%S 0,0,1,0,2,1,0,3,2,2,0,4,3,4,2,0,5,4,6,4,3,0,6,5,8,6,6,3,0,7,6,10,8,9,
%T 6,4,0,8,7,12,10,12,9,8,4,0,9,8,14,12,15,12,12,8,5,0,10,9,16,14,18,15,
%U 16,12,10,5,0,11,10,18,16,21,18,20,16,15,10,6
%N Triangle T read by rows: T(n, k) = (n - k)*(1 - (-1)^k + 2*k)/4, with 0 <= k < n.
%C T(n, k) is the k-th super- and subdiagonal sum of the matrix M(n) whose permanent is A332566(n).
%C The h-th subdiagonal of the triangle T gives 0 followed by the multiples of h+1 repeated.
%C For k > 0, the (2*k-1)-th and (2*k)-th columns of the triangle T give the multiples of k.
%F O.g.f.: y*(x*(2 + y + y^2) - (1 + y + 2*y^2))/((1 - x)^2*(1 - y)^3*(1 + y)^2).
%F T(n, k) = k*(n - k)/2 for k even.
%F T(n, k) = (1 + k)*(n - k)/2 for k odd.
%e n\k| 0 1 2 3 4 5
%e ---+------------
%e 1 | 0
%e 2 | 0 1
%e 3 | 0 2 1
%e 4 | 0 3 2 2
%e 5 | 0 4 3 4 2
%e 6 | 0 5 4 6 4 3
%e ...
%e For n = 4 the matrix M(4) is
%e 0 1 1 2
%e 1 0 1 1
%e 1 1 0 1
%e 2 1 1 0
%e and therefore T(4, 0) = 0, T(4, 1) = 3, T(4, 2) = 2 and T(4, 3) = 2.
%t T[n_,k_]:=(n-k)(1-(-1)^k+2k)/4; Flatten[Table[T[n,k],{n,1,12},{k,0,n-1}]] (* or *)
%t r[n_] := Table[SeriesCoefficient[y*(x*(2 + y + y^2) - (1 + y + 2*y^2))/((1 - x)^2 *(1 - y)^3 (1 + y)^2), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 0, n-1}]; Flatten[Array[r, 12]]
%Y Cf. A332566.
%Y Cf. A000004: 1st column; A000027: 2nd and 3rd column; A004526: diagonal; A005843: 4th and 5th column; A052928: 1st subdiagonal; A168237: 2nd subdiagonal; A168273: 3rd subdiagonal; A173196: row sums.
%K easy,nonn,tabl
%O 1,5
%A _Stefano Spezia_, Mar 08 2020
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