TITLE: Recognizing a measure whose moments are the motzkin numbers
QUESTION [9 upvotes]: The Catalan numbers are the moments of the Wigner semicircle distribution.
$$ \frac{1}{2\pi} \int_{-2}^2 x^{2n} \sqrt{4 - x^2} dx = \frac{1}{n+1} \binom{2n}{n} $$
Motzkin numbers enumerate the number of paths from (0,0) to (0,n) in steps of (1,1),(1,-1),(1,0) remaining above the x-axis.
For this particular sequence of numbers 1, 1, 2, 4, 9, 21, 51, 127... satisfying the recursion 
$$ M_n = \frac{3(n-1)M_{n-2} +(2n+1)M_{n-1}}{n+2} $$
Is it possible to reconstruct a measure $\rho(x)$ whose moments $\langle x^n \rangle$ follow that sequence?

REPLY [10 votes]: The formula for the measure of the Motzkin numbers as stated by Gjergij follows from the formula for the Catalan numbers if we write the formula for the Catalan numbers in the form
$\frac{1}{{2\pi }}\int_{ - 2}^2 {x^{2n} \sqrt {4 - x^2 } } dx = C_n $ and observe that the corresponding orthogonal polynomials are the Fibonacci polynomials  $F_{n + 1} (x, - 1)$  defined by $F_n (x, - 1) = xF_{n - 1} (x, - 1) - F_{n - 2} (x, - 1)$  with initial values $F_0 (x, - 1) = 0$ and $F_1 (x, - 1) = 1.$
For the Motzkin numbers the corresponding orthogonal polynomials are $F_{n + 1} (x - 1, - 1).$ Therefore a simple transformation of the integral gives the result.
Edit:
More generally (as answer to the comments by Gjergji and Brendan): 
Let $f(z) = \sum {r(n,c,d)z^n } $ satisfy $f(z) = 1 + czf(z) + d^2 z^2 f(z)^2. $
Then 
$r(n,c,d) = \frac{1}{{2\pi d^2 }}\int_{c – 2d}^{c + 2d} {x^n \sqrt {4d^2  - (x - c)^2 } } dx.$