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A395765
First column of the triangular array with T(0, m) = m! and T(n, m) = T(n - 1, m + 3) - T(n - 1, m + 2).
1
1, 4, 504, 256320, 339696000, 929459059200, 4557208289356800, 36403110891295948800, 442406862895153508352000, 7768652131253698010726400000, 189329460672170539873521008640000, 6200149639791767548556682032578560000, 265694561569328384410252384995689103360000
OFFSET
0,2
COMMENTS
a(n) is the first entry in the n-th row of a triangular array built from the factorial sequence, where T(0, m) = m! and each subsequent row is obtained by T(n, m) = T(n - 1, m + 3) - T(n - 1, m + 2).
The rows are generated by iterating the operator E^2*Delta, where E is the forward shift operator, E(f(m)) = f(m + 1), and Delta = E - 1 is the forward difference operator. Equivalently, E^2*Delta = E^2*(E - 1) = E^3 - E^2.
This is the (p, q) = (1, 2) case of the family obtained by iterating E^q*Delta^p on the factorial sequence. The case (p, q) = (1, 0) gives the derangement numbers A000166 and the case (p, q) = (1, 1) gives A033815.
FORMULA
Let f(m) = m!. Then:
a(n) = ((E^2*Delta)^n f)(0).
a(n) = Sum_{j=0..n} (-1)^j*binomial(n, j)*(3*n - j)!.
a(n) = Integral_{t=0..oo} t^(2*n)*(t - 1)^n*exp(-t) dt.
More generally, if b_{p, q}(n) = ((E^q*Delta^p)^n f)(0), where f(m) = m!, then b_{p, q}(n) = Integral_{t=0..oo} t^(q*n)*(t - 1)^(p*n)*exp(-t) dt = Sum_{j=0..p*n} (-1)^j*binomial(p*n, j)*((p + q)*n - j)!.
a(n) ~ sqrt(2*Pi) * 3^(3*n + 1/2) * n^(3*n + 1/2) / exp(3*n + 1/3). - Vaclav Kotesovec, May 13 2026
EXAMPLE
First column of
row 0: 1 1 2 6 24 120 720 5040
row 1: 4 18 96 600 4320 35280 322560
row 2: 504 3720 30960 287280 2943360 33022080
row 3: 256320 2656080 30078720 369774720 4906137600
row 4: 339696000 4536362880 64988179200 994447238400
MATHEMATICA
Table[Sum[(-1)^j * Binomial[n, j] * (3*n - j)!, {j, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 13 2026 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Dalton Heilig, May 05 2026
STATUS
approved