OFFSET
0,6
FORMULA
T(n,k) = [m^k] n! Sum_{s=0..n} Sum_{t=0..n} binomial(m,s) * binomial(m-s,t) * Sum_{i=0..s} Sum_{j=0..t} (-1)^(i+j) * binomial(s,i) * binomial(t,j) * binomial((s-i)*(t-j), n).
A052296(m,n) = Sum_{k=0..2*n} T(n,k) m^k / n!.
A092552(n) = -Sum_{i=1..n} S2(n,i)*T(i,1).
T(n,k) = Sum_{i=k/2..n} Sum_{s=0..i} Sum_{t=0..i} (-1)^(n-k-i+s+t) * S1(n, i) * S1(s+t, k) * S2(i, s) * S2(i, t), where S1(n,k) = A132393 and S2(n,k) = A008277 are the Stirling numbers of the first and second kind, respectively. - Eytan Chong, Feb 11 2026
EXAMPLE
Triangle begins:
0;
0, -1, 1;
0, -2, 5, -4, 1;
0, -24, 62, -63, 33, -9, 1;
0, -516, 1522, -1830, 1195, -470, 114, -16, 1;
The n = 3 row corresponds to the polynomial (-24*m + 62*m^2 - 63*m^3 + 33*m^4 - 9*m^5 + m^6)/3!, which counts the number of labeled digraphs with m nodes and 3 arcs and without directed paths of length >= 2.
MATHEMATICA
T[n_, k_]:= SeriesCoefficient[ n! Sum[Sum[Binomial[m, s]Binomial[m-s, t] Sum[Sum[ (-1)^(i+j) Binomial[s, i]Binomial[t, j]Binomial[(s-i)(t-j), n], {j, 0, t}], {i, 0, s}], {t, 0, n}], {s, 0, n}], {m, 0, k}]; Table[T[n, k], {n, 5}, {k, 0, 2n}]//Flatten (* Stefano Spezia, Oct 04 2025 *)
PROG
(Python)
import sympy as sp
from sympy.functions.combinatorial import numbers
def T(n, k):
sum = 0
for i in range(sp.ceiling(k/2), n+1):
for s in range(0, i+1):
for t in range(0, i+1):
if (s + t >= k):
sum += numbers.stirling(n, i, signed=True) * numbers.stirling(s+t, k, signed=True) * numbers.stirling(i, s) * numbers.stirling(i, t)
return sum # Eytan Chong, Feb 12 2026
CROSSREFS
KEYWORD
tabf,sign
AUTHOR
Eytan Chong, Oct 04 2025
STATUS
approved
