We might call such numbers "strictly bouncy numbers"; they exclude most ndigit "bouncy numbers" (cf. A152054) for n >= 4.
As n increases, a(n) approaches c/(2*cos(Pi*9/19))^n,
where c is 2.32290643963522604128193759601...
Is c the result of some simple expression?
From Jon E. Schoenfield, Dec 16 2008: (Start)
We could define the recursive formula
f(n) = 5*f(n1) + 10*f(n2)  20*f(n3)  15*f(n4) + 21*f(n5) + 7*f(n6)  8*f(n7)  f(n8) + f(n9)
and use a(n)=f(n) for n > 2 (a(n)=0 otherwise). Working backwards, given the terms f(11)=a(11) down through f(3)=a(3), the recursive formula would yield f(2)=81, f(1)=17 and f(0)=1, followed by the values 2, 1, 2, 2, 4, 5, 10, 14, 28, 42, 84, 132, etc., for negative values of n; these values are negative Catalan numbers for even n and twice (positive) Catalan numbers for odd n, down to f(16).
The above results apply for numbers in base 10. In general, for base m+1 (so that the largest possible value for a digit is m), we can write
a(n) = f(n) for n > 2, 0 otherwise, where
f(n) = Sum_{j=1..m} (1)^floor((j1)/2)*binomial(floor((m+j)/2),j)*f(nj) for n > 2,
f(2) = m^2, f(1) = 2*m  1, f(0)=1,
f(n) = 2*Catalan((1n)/2) for odd n, 2  2m < n < 0 and
f(n) = Catalan(n/2) for even n, 2  2m <= n < 0.
(The expressions for n < 0 work more than far enough down to give enough terms to begin generating f(3), f(4), etc.) (End)
