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

%I A091752 #42 Jul 02 2022 17:01:49
%S A091752 1,2,-2,1,40,-40,20,-6,1,2240,-2240,1120,-360,80,-12,1,246400,-246400,
%T A091752 123200,-40320,9520,-1680,220,-20,1,44844800,-44844800,22422400,
%U A091752 -7392000,1786400,-332640,48720,-5600,490,-30,1,12197785600,-12197785600,6098892800,-2018016000,493292800
%N A091752 Generalized Stirling2 array (-1,2)S2. Irregular triangle a(n, m) for n >= 1 and 2 <= m <= 2*n.
%C A091752 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.
%C A091752 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.
%C A091752 The sequence of row lengths for this array is [1,3,5,7,9,11,...] = A005408(n-1), n >= 1.
%C A091752 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
%C A091752 From _Wolfdieter Lang_ and _Werner Schulte_, Jan 29 2020: (Start)
%C A091752 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.
%C A091752 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.
%C A091752 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.
%C A091752 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.
%C A091752 The present entry is the instance s = 2, with the formulas given below. (End)
%H A091752 Leonard Carlitz, <a href="https://www.jstor.org/stable/2371100">On Arrays of Numbers</a>, Am. J. Math., 54,4 (1932) 739-752.
%H A091752 Wolfdieter Lang, <a href="/A091752/a091752.txt">First 6 rows</a>.
%F A091752 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.
%F A091752 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).
%F A091752 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
%F A091752 From _Werner Schulte_, Jan 29 2020: (Start)
%F A091752 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.
%F A091752 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).
%F A091752 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).
%F A091752 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.
%F A091752 (End)
%F A091752 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
%e A091752 Triangle starts:
%e A091752 {1},
%e A091752 {2, -2, 1},
%e A091752 {40, -40, 20, -6, 1},
%e A091752 {2240, -2240, 1120, -360, 80, -12, 1},
%e A091752 {246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1}.
%t A091752 w[n_, k_] := (Gamma[n-k/3] Gamma[1/3+n-k/3])/(Gamma[1/3-k/3] Gamma[-k/3]);
%t A091752 T[n_, k_] := 9^n Sum[(-1)^(k - j)  w[n, j]/((k - j)! j!), {j, 1, k}];
%t A091752 Table[Round[T[n,k]], {n,1,6}, {k, 2, 2 n}] (* _Peter Luschny_, Feb 05 2020 *)
%Y A091752 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).
%Y A091752 Cf. A078740, A331816.
%K A091752 sign,easy,tabf
%O A091752 1,2
%A A091752 _Wolfdieter Lang_, Feb 27 2004