G.f.: [1-tz-sqrt(1-2tz+t^2*z^2-4z^2)]/(2z^2).
T(n, k) = n!/[k!((n-k)/2)!((n-k)/2-1)! ] =
A055151(n, (n-k)/2) if n-k is a nonnegative even number; otherwise T(n, k) = 0.
T(n, k) = C(n, k)*C((n-k)/2)*(1+(-1)^(n-k))/2 if k <= n, 0 otherwise. -
Paul Barry, May 18 2005
G.f.: 1/(1-x*y-x^2/(1-x*y-x^2/(1-x*y-x^2/.... (continued fraction). -
Paul Barry, Dec 15 2008
E.g.f.: M(x,t) = e^(xt) AC(t) = e^(xt) I_1(2t)/t = e(xt) * e.g.f.(
A126120(t)) = e^(xt) Sum_{n>=0} t^(2n)/(n!(n+1)!) = exp[t P(.,x)].
The e.g.f. of this Appell sequence of polynomials P(n,x) is e^(xt) times the e.g.f. AC(t) of the aerated Catalan numbers
A126120. AC(t) = I_1(2t)/t, where I_n(x) = T_n(d/dx) I_0(x) are the modified Bessel functions of the first kind and T_n, the Chebyshev polynomials of the first kind.
P(n,x) has the lowering and raising operators L = d/dx = D and R = d/dD log{M(x,D)} = x + d/dD log{AC(D)} = x + Sum_{n>=0} c(n) D^(2n+1)/(2n+1)! with c(n) = (-1)^n
A180874(n+1), i.e., L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
(P(.,x) + y)^n = P(n,x+y) = Sum_{k=0..n} binomial(n,k) P(k,x) y^(n-k) = (b.+x+y)^n, where (b.)^k = b_k =
A126120(k).
Exp(b.D) e^(xt) = exp[(x+b.)t] = exp[P(.,x)t] = e^(b.t) e^(xt) = e^(xt) AC(t).
See p. 12 of the Alexeev et al. link and
A055151 for a refinement.
Shifted o.g.f: G(x,t) = [1-tx-sqrt[(1-tx)^2-4x^2]] / 2x = x + t x^2 + (1+t) x^3 + ... has the compositional inverse Ginv(x,t) = x / [1 + tx + x^2] = x - t x^2 +(-1+t^2) x^3 + (2t-t^3) x^4 + (1-3t^2+t^4) x^5 + ..., a shifted o.g.f. for the signed Chebyshev polynomials of the second kind of
A049310 (cf. also the Fibonacci polynomials of
A011973). Then the inversion formula of
A134264, involving non-crossing partitions and free probability with their multitude of interpretations (cf.
A125181 also), can be used with h_0 = 1, h_1 = t, and h_2 = 1 to interpret the coefficients of the Motzkin polynomials combinatorially.
(End)
Provides coefficients of the inverse of f(x) = x / [1 + h1 x + h2 x^2], a bivariate generating function of
A049310 (mod signs).
finv(x) = [(1-h1*x) - sqrt[(1-h1*x)^2-4h2*x^2]]/(2*h2*x) = x + h1 x^2 + (h2 + h1^2) x^3 + (3 h1 h2 + h1^3) x^4 + ... is a bivariate o.g.f. for this entry.
The infinitesimal generator for finv(x) is g(x) d/dx with g(x) = 1 /[df(x)/dx] = x^2 / [(f(x))^2 (1 - h2 x^2)] = (1 + h1 x + h2 x^2)^2 / (1 - h2 x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomials FI(n,h1,h2) of the compositional inverse of f(x), i.e., exp[x g(u)d/du] u |_(u=0) = finv(x) = 1 / [1 -x FI(.,h1,h2)]. Cf.
A145271. E.g.,
FI(0,h1,h2) = 0
FI(1,h1,h2) = 1
FI(2,h1,h2) = 1 h1
FI(3,h1,h2) = 1 h2 + 1 h1^2
FI(4,h1,h2) = 3 h2 h1 + 1 h1^3
FI(5,h1,h2) = 2 h2^2 + 6 h2 h1^2 + 1 h1^4
FI(6,h1,h2) = 10 h2^2 h1 + 10 h2 h1^3 + 1 h1^5.
And with D = d/dh1, FI(n+1, h1,h2) = MT(n,h1,h2) = (b.y + h1)^n = Sum_{k=0..n} binomial(n,k) b(k) y^k h1^(n-k) = exp[(b.y D] (h1)^n = AC(y D) (h1)^n, where b(k) =
A126120(k), y = sqrt(h2), and AC(t) is defined in my Jan 23 formulas above. Equivalently, AC(y D) e^(x h1) = exp[x MT(.,h1,h2)].
The MT polynomials comprise an Appell sequence in h1 with e.g.f. e^(h1*x) AC(xy) = exp[x MT(.,h1,h2)] with lowering operator L = d/dh1 = D, i.e., L MT(n,h1,h2) = dMT(n,h1,h2)/dh1 = n MT(n-1,h1,h2) and raising operator R = h1 + dlog{AC(y L)}/dL = h1 + Sum_{n>=0} c(n) h2^(n+1) D^(2n+1)/(2n+1)! = h1 + h2 d/dh1 - h2^2 (d/dh1)^3/3! + 5 h2^3 (d/dh1)^5/5! - ... with c(n) = (-1)^n
A180874(n+1) (consistent with the raising operator in the Jan 23 formulas).
The compositional inverse finv(x) is also obtained from the non-crossing partitions of
A134264 (or
A125181) with h_0 = 1, h_1 = h1, h_2 = h2, and h_n = 0 for all other n.
See
A238390 for the umbral compositional inverse in h1 of MT(n,h1,h2) and inverse matrix.
(End)
z1(x,h1,h2) = finv(x), the bivariate o.g.f. above for this entry, is the zero that vanishes for x=0 for the quadratic polynomial Q(z;z1(x,h1,h2),z2(x,h1,h2)) = (z-z1)(z-z2) = z^2 - (z1+z2) z + (z1*z2) = z^2 - e1 z + e2 = z^2 - [(1-h1*x)/(h2*x)] z + 1/h2, where e1 and e2 are the elementary symmetric polynomials for two indeterminates.
The other zero is given by z2(x,h1,h2) = (1 - h1*x)/(h2*x) - z1(x,h1,h2) = [1 - h1*x + sqrt[(1-h1*x)^2 - 4 h2*x^2]] / (2h2*x).
The two are the nontrivial zeros of the elliptic curve in Legendre normal form y^2 = z (z-z1)(z-z2), (see Landweber et al., p. 14, Ellingsrud, and
A121448), and the zeros for the Riccati equation z' = (z - z1)(z - z2), associated to soliton solutions of the KdV equation (see Copeland link).
(End)
Comparing the shifted o.g.f. S(x) = x / (1 - h_1 x + h_2 x^2) for the bivariate Chebyshev polynomials S_n(h_1,h_2) of
A049310 with the shifted o.g.f. H(x) = x / ((1 - a x)(1 - b x)) for the complete homogeneous symmetric polynomials H_n(a,b) = (a^(n+1)-b^(n+1)) / (a - b) shows that S_n(h_1,h_2) = H_n(a,b) for h_1 = a + b and h_2 = ab and, conversely, a = (h_1 + sqrt(h_1^2 - 4 h_2)) / 2 and b = (h_1 - sqrt(h_1^2 - 4 h_2)) / 2. The compositional inverse about the origin of S(x) gives a bivariate o.g.f. for signed Motzkin polynomials M_n(h_1,h_2) of this entry, and that of H(x) gives one for signed Narayana polynomials N_n(a,b) of
A001263, thereby relating the bivariate Motzkin and Narayana polynomials by the indeterminate transformations. E.g., M_2(h_1,h_2) = h_2 + h_1^2 = ab + (a + b)^2 = a^2 + 3 ab + b^2 = N_2(a,b). -
Tom Copeland, Jan 27 2024
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