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Is there a good closed form expression for the generating function of the formal power series $$ A(z) := \sum_{n=0}^\infty z^{2^n} = z + z^2 + z^4 + z^8 + z^{16} + \cdots. $$ Is there a tractable way to retrieve the coefficient of $z^m$ in powers of $A(z)$, say in $A(z)^k$ for $k \geq 1$? Thanks.

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    $\begingroup$ Not a closed form expression, but there is a continued fraction expression: $x / (1 - x / (1 + x / (1 + x / (1 - x / (1 + x / (1 - x / \ldots)))))$ See oeis.org/A209229 $\endgroup$ Dec 23, 2015 at 22:11
  • $\begingroup$ How could we use this to retrieve the coefficient of $z^m$ in $A(z)^k$, i.e., in powers of $A(z)$? $\endgroup$
    – user152169
    Dec 23, 2015 at 22:47
  • $\begingroup$ Almost certainly no. It is already highly unlikely that there is a closed form for $A(1/2)$. $\endgroup$ Dec 23, 2015 at 23:17

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The value $A(1/2)=\kappa$ is known as the Kempner number, and was proven transcendental in 1916. The paper "The Many Faces of the Kempner Number", by Adamczewski, may provide some insight for you.

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How about $A(z)=\frac{z^2}{1-z^2}$? This works if $|z|<1$.

I got this idea from expanding $\frac{1}{1-z}$.

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    $\begingroup$ No, you’re getting the even powers; the OP wants the exponents to be powers of $2$. $\endgroup$ Dec 23, 2015 at 22:12
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    $\begingroup$ Oh I see... my fault. $\endgroup$
    – user300627
    Dec 23, 2015 at 22:14
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    $\begingroup$ You can remove your answer by clicking on the delete option at the bottom of your post. $\endgroup$
    – C. Falcon
    Dec 23, 2015 at 22:17

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