A330613 Triangle read by rows: T(n, k) = 1 + k - 2*n - 2*k*n + 2*n^2, with 0 <= k < n.

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, 33, 20, 7, 113, 98, 83, 68, 53, 38, 23, 8, 145, 128, 111, 94, 77, 60, 43, 26, 9, 181, 162, 143, 124, 105, 86, 67, 48, 29, 10, 221, 200, 179, 158, 137, 116, 95, 74, 53, 32, 11

Offset

1,2

Comments

T(n, k) is the k-th super- and subdiagonal sum of the matrix M(n) whose permanent is A330287(n).

Links

Stefano Spezia, First 150 rows of the triangle, flattened

Formula

O.g.f.: (1 - x*(2 - 5*x + 2*(1 + x)*y))/((1 - x)^3*(1 - y)^2).

E.g.f.: exp(x+y)*(1 + 2*x*(x - y) + y).

T(n, k) = A001844(n-1) - k*A005408(n-1), with 0 <= k < n. [Typo corrected by Stefano Spezia, Feb 14 2020]

Example

n\k|   0   1   2   3   4   5
---+------------------------
1  |   1
2  |   5   2
3  |  13   8   3
4  |  25  18  11   4
5  |  41  32  23  14   5
6  |  61  50  39  28  17   6
...
For n = 3 the matrix M is
      1, 2, 3
      2, 4, 6
      3, 6, 8
and therefore T(3, 0) = 1 + 4 + 8 = 13, T(3, 1) = 2 + 6 = 8 and T(3, 2) = 3.

Mathematica

Flatten[Table[1+k-2n-2k*n+2n^2, {n, 1, 11}, {k, 0, n-1}]] (* or *)
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 *)
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]]

Crossrefs

Cf. A005408, A330287.

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.

Sequence in context: A294684 A013946 A261327 * A085436 A277710 A286148

Adjacent sequences: A330610 A330611 A330612 * A330614 A330615 A330616

Keyword

easy,nonn,tabl

Author

Stefano Spezia, Dec 20 2019

Status

approved