A333119 Triangle T read by rows: T(n, k) = (n - k)*(1 - (-1)^k + 2*k)/4, with 0 <= k < n.

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, 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, 16, 12, 10, 5, 0, 11, 10, 18, 16, 21, 18, 20, 16, 15, 10, 6

Offset

1,5

Comments

T(n, k) is the k-th super- and subdiagonal sum of the matrix M(n) whose permanent is A332566(n).

The h-th subdiagonal of the triangle T gives 0 followed by the multiples of h+1 repeated.

For k > 0, the (2*k-1)-th and (2*k)-th columns of the triangle T give the multiples of k.

Links

Table of n, a(n) for n=1..78.

Formula

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

T(n, k) = k*(n - k)/2 for k even.

T(n, k) = (1 + k)*(n - k)/2 for k odd.

Example

n\k| 0 1 2 3 4 5
---+------------
1  | 0
2  | 0 1
3  | 0 2 1
4  | 0 3 2 2
5  | 0 4 3 4 2
6  | 0 5 4 6 4 3
...
For n = 4 the matrix M(4) is
      0 1 1 2
      1 0 1 1
      1 1 0 1
      2 1 1 0
and therefore T(4, 0) = 0, T(4, 1) = 3, T(4, 2) = 2 and T(4, 3) = 2.

Mathematica

T[n_, k_]:=(n-k)(1-(-1)^k+2k)/4; Flatten[Table[T[n, k], {n, 1, 12}, {k, 0, n-1}]] (* or *)
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]]

Crossrefs

Cf. A332566.

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.

Sequence in context: A063942 A263405 A106384 * A356582 A320839 A094314

Adjacent sequences: A333116 A333117 A333118 * A333120 A333121 A333122

Keyword

easy,nonn,tabl

Author

Stefano Spezia, Mar 08 2020

Status

approved