Interpolation of a sequence of polynomials (viewed in terms of q-analogue powers)

I'm trying to find closed-form expressions for a sequence of coefficients, such that the index of the coefficient occurs as number such that I can later interpolate to fractional indexes as well. Surely this is in most cases not unique, but if the occurence is a very simple function of the index then a "natural" interpolation is worth a try.

I managed so far to express that coefficents in terms of q-analogues (with the notation $\small [n]_q = {q^n-1 \over q-1 }$ ) and the index h occurs as a simple number, unfortunately not strictly increasing.

For one of my "simpler" sequences I managed to get the following pattern:

$\qquad \qquad \small \begin{array} {rll} h & &\text{entry} \\\ 1 &: &[1]_q \cdot [0]_{q^2} \\\ 2 &:&[1]_q \cdot [1]_{q^2} \\\ 3 &:&[3]_q \cdot [1]_{q^2} \\\ 4 &:&[3]_q \cdot [2]_{q^2} \\\ 5 &:&[5]_q \cdot [2]_{q^2} \\\ 6 &:&[5]_q \cdot [3]_{q^2} \\\ \ldots & \ldots \end{array}$

and my question is, whether there might exist some reformulation of the expressions which I do not see, where I can get a strictly increasing simple continuous function of the index h . For instance I tried to express the coefficients in different factorizations and as sums, but not yet a suggestive/"natural" one.

For instance entry at h=4 is originally $\small q^4+q^3+2 q^2+q^1 + 1$ and I decided to factorize this way $\small q^4+q^3+2 q^2+q^1 + 1 =(q^2+q^1+1)*(q^2+1) = [3]_q \cdot [2]_{q^2}$ because that looked somehow obvious also in the context of the other entries.
But that factorizing is not unique, and also one could separate this into finite sums first to get another pattern. Because that all is somehow arbitrary I think this is useful only if there is some "obvious" re-expression which I don't recognize but someone with experience with q-analogues has an eye for this.

Context: the problem stems from analysis of iteration of functions $f(x)=ax+bx^2+cx^3+...$ in symbolic representation where h indicates the h'th iteration/self-composition.

(I'm not convinced that I've taken the best selection of "tags" for this question, sorry)

[update] Here is a table for the factoring of the first few coefficients if a function $f(x)=ax+bx^2+cx^3+\ldots$ is h -times iterated:

[update2] The procedure of diagonalization can be done even symbolically and gives polynomials for the coefficients at each power of x in terms of h. I show the examples here for the cofactors $\small C_k(h)$ of $\small x^k \cdot b^{k-1}$ where Andrew for instance gave a q-binomial-decomposition for the cofactor $\small C_4(h)$. For $\small C_3(h)$ diagonalization gives

$$C_3(h) = \small {N_3(h) \over D_3(h)} = \small {\begin{bmatrix} -( &2 a^{2h} &- 2 a^{1h}&) & \cdot a^1 \\ +( &2 a^{3h} &- 2 a^{2h}&) & \cdot a^0 \end{bmatrix} \over a^2 (a-1)^2 \operatorname{qfac} (2,a) }$$ (where $\small \operatorname{qfac}(n,a) = [1]_a \cdot [2]_a \cdot [3]_a \ldots \cdot [n]_a$ and the misuse of the matrix-notation is just for readability/convenience: we evaluate just the sum of the entries ) which boils down to $$C_3(h) = \small { 2 \cdot a^h \cdot (a^h-1) \cdot (a^h-a) \over a^2 (a-1)^2 \operatorname{qfac} (2,a) } = 2a^{h-1} {\operatorname{qfac} (h,a) \over \operatorname{qfac} (h-2,a) \operatorname{qfac} (2,a) } = 2a^{h-1} [h:2]_a$$ and allows the expression by the q-binomial (here $\small [m:n]_q$ is the q-analogue to binomial(m,n) ) , at least for integer h .

For the coefficient $\small C_4(h)$ at $\small x^4 b^3$ diagonalization gives $$C_4(h)= \small {N_4(h) \over D_4(h)} = \small { \begin{bmatrix} +(& & &+ 5 a^{2h} &- 5 a^h & ) \cdot a^3 \\ + (& & - 6 a^{3h} &+ 6 a^{2h} & & ) \cdot a^2 \\ + (&+ 1 a^{4h}& - 6 a^{3h} &+ 6 a^{2h} &- 1 a^h & ) \cdot a \\ + (&+ 5 a^{4h}& - 6 a^{3h} &+ 1 a^{2h} & & ) \end{bmatrix} \over a^3 (a-1)^3 \operatorname{qfac} (3,a) }$$

and I confess, that I'm beginning to doubt, whether the decomposition into q-binomials can lead to a smoother representation for coefficients at higher powers of x at all...

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Sorry I'm not quite getting it. You are looking for sequences $h$ that have closed form expressions? Is there some other special property you are looking for that your example satisfies? – John M Aug 17 '11 at 13:22
@John: If I re-expand the expressions under "entry" then I get certain polynomials in powers of q. The sequence of that polynomials is my original source. By inspection of the exponents I felt, that the above factorization was the most "concise", where also the index h appears in each polynomial/entry by a certain formula, but not exactly unique and not strictly increasing. I'll add some more explanation in the question above – Gottfried Helms Aug 17 '11 at 15:12
It looks like you've got an odd-even pattern going on in your sequence, and it's unlikely to yield an interpolable expression because it very likely inherently requires step functions. Your best hope may be to look for q-function identities that might 'combine' the floor terms into something that wouldn't involve the floor functions. – Steven Stadnicki Aug 17 '11 at 15:59
Incidentally, do you have any reason to expect a closed form for your results? Fractional iteration is notorious for yielding functions without a closed form (for instance, the classic $f(f(x)) = e^x$ problem). A more typical approach for these problems is to find a suitable conjugate function $g$ such that (for instance) $f(g(x)) = g(cx)$; then $f^{(n)}(x) = g(c^ng^{-1}(x))$, and now $n$ is interpolable. The known solvable quadratic iterations all yield to this method, for instance. – Steven Stadnicki Aug 17 '11 at 16:03
@Gottfried the coefficient at x^4 b^3 seems to be $5 a^2 {{h}\brack{3}}_a+{{h+1}\brack{3}}_a$, I checked it up to $h=8$. – Andrew Aug 18 '11 at 9:53

It's the q-binomial coefficients ${h\brack 2}_q$.