# Is there a reasonably simple, non-recursive formula for this sequence?

I have been curious about the following sequence of rational numbers for some time. I identified the numerator and denominators on the Online Encyclopedia of Integer sequences, but there was not much information there, except for an alternate way of calculating the sequence, using series reversions. I've verified that that works, though I don't quite understand why it works.

$A_n$ = A097088 / 2 ^ A097088

$a_1$ = 1, $a_2$ = 1/2

If f is the formal series for A (with $a_0$ = 0, so no constant term), then one recurrence formula is:

$a_{n+1}$ = - ([$x^{n+1}$] ($a_2$ $f^2$ + $a_3$ $f^3$ + ... $a_n$ ${f}^n$ )) / 2, for n >= 2

This recurrence only depends on the terms of f up to $a_n$, because f has no constant term.

Also, f(f(x)) = x + $x^2$, and the values of a can also be defined by using the chain rule repeatedly since f(0)=0.

$a_n$ = 1, $\frac{1}{2}$, -$\frac{1}{4}$, $\frac{1}{4}$, -$\frac{5}{16}$, $\frac{27}{64}$, -$\frac{9}{16}$, $\frac{171}{256}$ ...

$a_n$ has both negative and positive terms alternating without any clear pattern that I've seen, and I've calculated about 50 terms using Mathematica, and their absolute magnitude generally grows over time, but I haven't seen any way to get to a decent non-recursive formula, and maybe there is no such formula. But I am curious if anyone else has run into this type of sequence and knows of tricks for finding a formula.

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It may not help, but these are related to A027436 $f(f(x)) = x(1 + 4x)$ in the sense that $a_n =\text{A027436}_n / 4^{n-1}$ and that might give you an integer sequence to work with rather than fractions. – Henry Apr 20 '11 at 18:54
Just for your initial formula, the calculations are unclear to me. If A097088(5) = 5, then $A_5$ by your definition $= 5\cdot 32 = 180$ and $A_6 = \frac{27}{2^27}$. What is the relation between $A_n$ and $a_n$? – Mitch Apr 20 '11 at 19:27