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%I #22 Feb 15 2020 02:03:50
%S 1,5,2,13,8,3,25,18,11,4,41,32,23,14,5,61,50,39,28,17,6,85,72,59,46,
%T 33,20,7,113,98,83,68,53,38,23,8,145,128,111,94,77,60,43,26,9,181,162,
%U 143,124,105,86,67,48,29,10,221,200,179,158,137,116,95,74,53,32,11
%N Triangle read by rows: T(n, k) = 1 + k - 2*n - 2*k*n + 2*n^2, with 0 <= k < n.
%C T(n, k) is the k-th super- and subdiagonal sum of the matrix M(n) whose permanent is A330287(n).
%H Stefano Spezia, <a href="/A330613/b330613.txt">First 150 rows of the triangle, flattened</a>
%F O.g.f.: (1 - x*(2 - 5*x + 2*(1 + x)*y))/((1 - x)^3*(1 - y)^2).
%F E.g.f.: exp(x+y)*(1 + 2*x*(x - y) + y).
%F T(n, k) = A001844(n-1) - k*A005408(n-1), with 0 <= k < n. [Typo corrected by _Stefano Spezia_, Feb 14 2020]
%e n\k| 0 1 2 3 4 5
%e ---+------------------------
%e 1 | 1
%e 2 | 5 2
%e 3 | 13 8 3
%e 4 | 25 18 11 4
%e 5 | 41 32 23 14 5
%e 6 | 61 50 39 28 17 6
%e ...
%e For n = 3 the matrix M is
%e 1, 2, 3
%e 2, 4, 6
%e 3, 6, 8
%e and therefore T(3, 0) = 1 + 4 + 8 = 13, T(3, 1) = 2 + 6 = 8 and T(3, 2) = 3.
%t Flatten[Table[1+k-2n-2k*n+2n^2,{n,1,11},{k,0,n-1}]] (* or *)
%t r[n_] := Table[SeriesCoefficient[(1-x*(2-5x+2(1+x)y))/((1-x)^3*(1-y)^2), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 0, n-1}]; Flatten[Array[r, 11]] (* or *)
%t r[n_] := Table[SeriesCoefficient[Exp[x+y]*(1+2x(x-y)+y), {x, 0, i}, {y, 0, j}]*i!*j!, {i, n, n}, {j, 0, n-1}]; Flatten[Array[r, 11]]
%Y Cf. A005408, A330287.
%Y Cf. A000027: diagonal; A001105: 2nd column; A001844: 1st column; A016789: 1st subdiagonal; A016885: 2nd subdiagonal; A017029: 3rd subdiagonal; A017221: 4th subdiagonal; A017461: 5th subdiagonal; A081436: row sums; A132209: 3rd column; A164284: 7th subdiagonal; A269044: 6th subdiagonal.
%K easy,nonn,tabl
%O 1,2
%A _Stefano Spezia_, Dec 20 2019
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