A395762 Array read by ascending antidiagonals: A(n, k) = Sum_{j=0..n} Stirling2(n, j)*5^j*(k)_j.

1, 0, 1, 0, 5, 1, 0, 5, 10, 1, 0, 5, 60, 15, 1, 0, 5, 160, 165, 20, 1, 0, 5, 360, 1215, 320, 25, 1, 0, 5, 760, 5565, 3920, 525, 30, 1, 0, 5, 1560, 21015, 35120, 9025, 780, 35, 1, 0, 5, 3160, 72165, 229520, 123525, 17280, 1085, 40, 1

Offset

0,5

Comments

Base 5 polynomial collapse array for the iterated inverse Pascal operator.

In general, for any integer base b >=1, A_b(n, k) = Sum_{j=0..n} Stirling2(n, j)*b^j*(k)_j. Thus, this entry is the case b=5. The cases b=2,3,4 correspond respectively to A394444, A395570, and A395572.

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^4(i^n*5^i)]_k = Sum_{j=0..n} Stirling2(n, j)*5^j*(k)_j.

For fixed k, the e.g.f. of A(n, k) in n is Sum_{n>=0} A(n, k)*u^n/n! = (1 + 5*(exp(u) - 1))^k.

A(n + 1, k) = k*(A(n, k) + 4*A(n, k - 1)) for k >= 1, with A(0, k) = 1 for k >= 0 and A(n, 0) = 0 for n > 0.

More generally, A_b(n, k) = Sum_{j=0..n} Stirling2(n, j)*b^j*(k)_j = Sum_{j=0..k} (-1)^(k - j)*binomial(k, j)*(b - 1)^(k - j)*b^j*j^n. For fixed k, the e.g.f. in n is Sum_{n>=0} A_b(n, k)*x^n/n! = (1 + b(e^x - 1))^k.

Example

k | 0   1    2      3       4        5
n |--------------------------------------
0 | 1   1    1      1       1        1
1 | 0   5   10     15      20       25
2 | 0   5   60    165     320      525
3 | 0   5  160   1215    3920     9025
4 | 0   5  360   5565   35120   123525
5 | 0   5  760  21015  229520  1320025
A(2,2) = Stirling2(2, 1)*5*(2)_1 + Stirling2(2, 2)*25*(2)_2 = 1*5*2 + 1*25*2 = 60.

Crossrefs

Cf. A394444, A395570, A395572.

Sequence in context: A019904 A002068 A021666 * A340841 A143148 A389454

Adjacent sequences: A395758 A395759 A395761 * A395763 A395764 A395765

Keyword

nonn,tabl

Author

Dalton Heilig, May 05 2026

Status

approved