A091752 Generalized Stirling2 array (-1,2)S2. Irregular triangle a(n, m) for n >= 1 and 2 <= m <= 2*n.

1, 2, -2, 1, 40, -40, 20, -6, 1, 2240, -2240, 1120, -360, 80, -12, 1, 246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1, 44844800, -44844800, 22422400, -7392000, 1786400, -332640, 48720, -5600, 490, -30, 1, 12197785600, -12197785600, 6098892800, -2018016000, 493292800

Offset: 1

Author

Wolfdieter Lang, Feb 27 2004

Keywords

• sign
• easy
• tabf

Comments

This a(n,m) array appears in the normal ordering formula ((1/x)*(d_x)^2)^n = Sum_{m=2..2*n} a(n,m)*x^(m-3*n)*(d_x)^m, n >= 1, with the derivative operator d_x := d/dx.

This is an extension of the generalized Stirling2 arrays S_{r,s}(n,k) (here k=m) considered for nonnegative r and s in the Blasiak et al. reference given in A078740. See also the Schork reference given there.

The sequence of row lengths for this array is [1,3,5,7,9,11,...] = A005408(n-1), n >= 1.

These generalized Stirling2 arrays have been treated in Carlitz's paper (with r = lambda + mu, s = mu), and a recurrence is given eq. (4). See the formula section for the present case mu = 2, lambda = -3, and Carlitz's a_{n,s-1} = a(n, s) (here s = m). - Wolfdieter Lang, Dec 16 2019

From Wolfdieter Lang and Werner Schulte, Jan 29 2020: (Start)

For the case of the (irregular) triangles (-1,s)S2, for s >= 1, W. Schulte conjectured that a(s; n, m) = T(s; (s+1)*n - m - 1, m - s), for n > = 1 and m = s, s+1, ..., s*n, with the row polynomials RT(s; n, x) = Sum_{m = 0..s*n} T(s; n, m)*x^m of the triangle T(s), n >= 0, defined by the Rodrigues-type formula exp(x^(s+1)/(s+1)) (d/dx)^n exp(-x^(s+1)/(s+1)) = (-1)^n*RT(s; n, x). Thus the rows of (-1,s)S2 are every (s+1)-th upwards antidiagonals of T(s), but with offset m = s instead of m = 0.

The proof of the conjecture follows by showing the Carlitz recurrence for (-1,s)S2, but with offset n = 0 and m = 0; that is, a(s; n+1, m+s) =: aHat(s; n, m) = Sum_{j = 0..s} binomial(s, j)*fallfac(s + m - j - (s+1)*n , s-j)*aHat(s; n-1, m-j), for n >= 0, m = 0, 1, ..., s*n, with aHat(s; 0, 0) = 1 and aHat(s; n, m) = 0 for m < 0 and m > s*n. The falling factorials are fallfac(x, n). With aHat(s; n, m) = T(s; (s+1)*n - m, m) this leads to a recurrence for T(s) which is equivalent to the recurrence for the row polynomials RT(s), namely RT(s; n, x) = Sum_{j=0 .. s} binomial(s, j)*x^j*fallfac(s - j - n, s - j)*RT(s; n - (s+1) + j, x), for n >= 1, and RT(s; 0, x) = 1. This, in turn, can be proved by induction over n >= 1 from the simpler recurrence for RT(s) obtained directly from the Rodrigues-type definition, namely RT(s; n, x) = x^s*RT(s; n-1, x) - (d/dx)RT(s; n-1, x), n >= 1,with RT(s; 0, x) = 1.

The e.g.f. of the triangle T(s), that is of the row polynomials {RT(s;n, x)}_{n>=0}, is E(s; t, x) = exp((x^(s+1) - (x - t)^(s+1))/(s+1)). This can be proved from the simple RT(s) recurrence, leading to (d/dt + d/dx)E(s; t, x) = x^s*E(s; t, x), with E(s; 0, x) = 1. After using E(s; t, x) = 1*exp(x^(s+1)/(s+1) + f(s; t, x)), with f(s; 0, x) = -x^(s+1)/(s+1), this becomes (d/dx - d/dt)f(s; t, x) = 0 meaning that f is a function of y = x - t, say, g(s; y) = -y^(s+1)/(s+1) because it has to become f(s; 0, x) for t = 0.

The explicit form for (-1,s)S2 is a(s; n, m) = (-1)^(n*s -m)*((s+1)*n - m-1)!/((s+1)^(n-1)*(n-1)!)*Sum_{j=0..floor((m-s)/(s+1))} (-1)^j* binomial(n-1, j)*binomial((s+1)*(n-1-j), m - s - (s+1)*j). One can prove the corresponding formula for T(s; n, m) by showing that it satisfies the T(s) recurrence T(s; n, m) = T(s; n-1, m-l) + (m+1)*T(s; n-1, m+1), for n >= 1, with T(s; 0, 0) = 1, and 0 for m < 0 or m > s*n.

The present entry is the instance s = 2, with the formulas given below. (End)

Examples

			Triangle starts:
{1},
{2, -2, 1},
{40, -40, 20, -6, 1},
{2240, -2240, 1120, -360, 80, -12, 1},
{246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1}.
		

Links

• Leonard Carlitz, On Arrays of Numbers, Am. J. Math., 54,4 (1932) 739-752.
• Wolfdieter Lang, First 6 rows.

Crossrefs

Column sequences for m=2..8 and n >= ceiling(m/2) are A052502(n-1), -A052502(n-1), A091535(n-1), -6*A091753(n), A091754(n), -12*A091755(n), A091756(n).

Cf. A078740, A331816.

Programs

Mathematica

w[n_, k_] := (Gamma[n-k/3] Gamma[1/3+n-k/3])/(Gamma[1/3-k/3] Gamma[-k/3]);
T[n_, k_] := 9^n Sum[(-1)^(k - j)  w[n, j]/((k - j)! j!), {j, 1, k}];
Table[Round[T[n,k]], {n,1,6}, {k, 2, 2 n}] (* Peter Luschny, Feb 05 2020 *)

Formula

a(n, m) = (((-1)^m)/m!)*Sum_{p=2..m} (-1)^p*binomial(m, p)*Product_{j=1..n} fallfac(p-3*(j-1), 2), n >= 1, 2 <= m <= 2*n, otherwise 0. From eq. (12) of the Blasiak et al. reference (see A078740) with r=-1, s=2, k=m.

Recurrence: a(n, m) = Sum_{p=0..2} binomial(2, p)*fallfac(-3*(n-1)+m-p, 2-p)*a(n-1, m-p), n >= 2, 2 <= m <= 2*n, a(1, 2) = 1, otherwise 0. Rewritten from eq. (19) of the Schork reference (see A078740) with r = -1, s = 2. fallfac(n, m) := A008279(n, m) (falling factorials triangle).

Recurrence (Carlitz): a(n m) = a(n-1, m-2) - 2*(3*n - (m +2))*a(n-1, m-1) + (3*n - (m + 3))*(3*n - (m + 2))*a(n-1, m), for n >= 2, m >= 1, and a(n, m) = 0 if m <= 1 or m > 2*n, and a(1, 2) = 1. - Wolfdieter Lang, Dec 16 2019

From Werner Schulte, Jan 29 2020: (Start)

a(n, m) = T(3*n - m - 1, m - 2), for n > = 1 and m = 2, 3, ..., 2*n, with the irregular triangle defined by (-1)^n*exp(x^3/3)*(d/dx)^n exp(-x^3/3) = RT(n, x) = Sum_{k=0..2*n} T(n, k)*x^k, for n >= 0. For T(n, k) see A331816.

The recurrence RT(n, x) = x^2*RT(n-1, x) - (d/dx)RT(n-1, x), n >= 1, with RT(0, x) = 1, implies the T recurrence T(n, k) = T(n-1, k-2) - (k+1)*T(n-1, k+1), for n >= 1, with T(0, 0) = 1, and T(n, m) = 0 for m < 0 and m > 2*n. Also, by induction over n: RT(n, x) = x^2*RT(n-1, x) - 2*(n-1)*x*RT(n-2, x) + (n-1)*(n-2)*RT(n-3, x), using the former recurrence and inserting the derivatives. This translates to an obvious further recurrence for the irregular triangle T. It is used in order to prove the Carlitz recurrence for the index shifted aHat(n, m) = a(n+1, m + 2).

The e.g.f. of the irregular triangle, that is of the row polynomials RT, is E(t, x) = exp((x^3 - (x - t)^3)/3). See the comment above for a proof (setting s=2 there).

The explicit form is a(n, m) = (-1)^m*(3*n - m - 1)!/(3^(n-1)*(n-1)!)*Sum_{j=0..floor((m-2)/3)} (-1)^j*binomial(n-1, j)*binomial(3*(n-1-j), m -2 - 3*j), for n >= 1, and 2 <= m <= 2*n.

(End)

T(n, k) = 9^n*Sum_{j=1..k} (-1)^(k-j)*w(n,j)/((k-j)!*j!) where w(n,k) = (Gamma(n-k/3)*Gamma((1-k)/3+n))/(Gamma((1-k)/3)*Gamma(-k/3)). - Peter Luschny, Feb 05 2020