OFFSET
1,2
COMMENTS
Coefficients of Sum_{k=1..n} T(n,k)*(m+k)!/((m-1)!*(n+1)!) = Sum_{i=1..m} i^n.
FORMULA
T(n,k) = (-1)^(n-k)*(n+1)!/(k+1)! * Sum_{i=0..k} (-1)^i*binomial(k,i)*(k-i)^n.
T(n+1,k+1) = (-1)^(n+k)*(n+2)*(k+1)*(abs(T(n,k)/(k+2)) + abs(T(n,k+1))) with T(n,1) = (-1)^(n+1)*(n+1)!/2 and T(n,k) = 0 if n < k.
EXAMPLE
The triangle T(n,k) begins:
n\k 1 2 3 4 5 6 7 8
1: 1
2: -3 2
3: 12 -24 6
4: -60 280 -180 24
5: 360 -3600 4500 -1440 120
6: -2520 52080 -113400 65520 -12600 720
7: 20160 -846720 3034080 -2822400 940800 -120960 5040
8: -181440 15361920 -87635520 123451776 -63504000 13789440 -1270080 40320
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MATHEMATICA
T[n_, k_] := (-1)^(n - k) * (n + 1)!/(k + 1) * StirlingS2[n, k]; Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Amiram Eldar, Sep 01 2022 *)
PROG
(Python)
def A(d):
A = [[0 for col in range(d)] for row in range(d)]
for i in range(d):
A[0][i] = 1
for i in range(1, d):
for j in range(i, d):
A[i][j] = (A[i][j - 1] + A[i - 1][j - 1])*(i + 1)
for i in range(d):
for j in range(i, d):
for k in range(i+3, j+3):
A[i][j] *= k
a = []
for i in range(d):
for j in range(i+1):
a.append((-1)**(i+j)*A[j][i])
return(a)
(PARI) T(n, k) = (-1)^(n-k)*(n+1)!*stirling(n, k, 2)/(k+1); \\ Michel Marcus, Sep 26 2022
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Samuel Gantner, Aug 31 2022
STATUS
approved