A395570 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} Stirling2(n, j)*3^j*(k)_j.

1, 0, 1, 0, 3, 1, 0, 3, 6, 1, 0, 3, 24, 9, 1, 0, 3, 60, 63, 12, 1, 0, 3, 132, 333, 120, 15, 1, 0, 3, 276, 1359, 984, 195, 18, 1, 0, 3, 564, 4869, 6600, 2175, 288, 21, 1, 0, 3, 1140, 16263, 37272, 20715, 4068, 399, 24, 1

Offset

0,5

Comments

Base 3 polynomial collapse array for the iterated inverse Pascal operator.

For base 2, the analogous one-step array is A394444.

Links

Table of n, a(n) for n=0..54.

Formula

Let D be the inverse Pascal operator: [D(f)]_m = Sum_{i=0..m} (-1)^(m-i)*binomial(m, i)*f(i). Then A(n, k) = [D^2(i^n*3^i)]_k = Sum_{j=0..n} Stirling2(n, j) *3^j*(k)_j.

Seen as a triangle, T(n, k) = A(n - k, k).

For fixed k, the e.g.f. of A(n, k) in n is Sum_{n>=0} A(n, k)*u^n/n! = (1 + 3*(exp(u) - 1))^k.

Example

  k |    0    1      2       3        4        5
  n |------------------------------------------------
  0 |    1    1      1       1        1        1
  1 |    0    3      6       9       12       15
  2 |    0    3     24      63      120      195
  3 |    0    3     60     333      984     2175
  4 |    0    3    132    1359     6600    20715
  5 |    0    3    276    4869    37272   169575
A(2, 2) = Stirling2(2, 1)*3*(2)_1 + Stirling2(2, 2)*9*(2)_2 = 24.

Crossrefs

Cf. A394444, A008277, A152751, A182464.

Sequence in context: A104556 A116089 A122016 * A067882 A215771 A379078

Adjacent sequences: A395567 A395568 A395569 * A395571 A395572 A395573

Keyword

nonn,tabl

Author

Dalton Heilig, Apr 28 2026

Status

approved