I've been exploring functions that have a general form:

$$\sum_{k=0}^\infty{ a^{b^k} } \tag{1}$$

In particular, I'm now checking this equality, which seems to hold:

$$2 \sum_{k=0}^\infty{ \left( \frac{1}{2^{2^k}} - \frac{1}{2^{2^k\cdot3-1}} \right) } = 5/6$$

I'm also in the process of finding more identities/equations, but I don't want to reinvent the wheel.

So I'm wondering, What is known about series of the form (1)? I'm interested in this and anything related to "doubly exponential" series. I'd be extremely interested in any books or papers that anyone knows about.

  • $\begingroup$ The first term of the series shown is $3/4$, so the expression is at least $3/2$. The second term is $9/32$, so even after excluding the first term the expression is at least $9/16 > 1/2$. $\endgroup$ Jan 13, 2014 at 20:02
  • $\begingroup$ @AntonioVargas: Oops, you're right. It should be $\frac{5}{6}$. I'll edit the question! $\endgroup$
    – Matt Groff
    Jan 13, 2014 at 20:03
  • $\begingroup$ But $5/6$ is still less than my first lower bound of $3/2$. Numerically the expression is about $2.196319819$. $\endgroup$ Jan 13, 2014 at 20:05
  • 1
    $\begingroup$ @AntonioVargas: Sorry, I'm still not sure if it's correct, but I edited the result again. It's my fault; I did a bunch of quick and dirty calculations, and I'm still working on checking and refining them. That's one of the reasons why I'm so interested in finding existing results. $\endgroup$
    – Matt Groff
    Jan 13, 2014 at 20:11
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    $\begingroup$ Your equality cannot possibly be true, since the binary expansion of the left-hand side is aperiodic, thus the sum is irrational. Actually, transcendental, by a Liouville-type argument. $\endgroup$ Jan 26, 2014 at 8:24

1 Answer 1


The function $$ f(z)=\sum_{k=0}^{\infty}z^{a^k}=z+z^a+z^{a^2}+z^{a^3}+\ldots, $$ where $a$ is a positive integer, is analytic for $|z|<1$, equal to $0$ at $z=0$, and satisfies the functional equation $$ f(z^a)=f(z)-z. $$ For $a=2$, you have the additional fun property that $$ f(z)+f(z^3)+f(z^5)+\ldots=\frac{z}{1-z}. $$

  • $\begingroup$ Functions of this form can have what are called natural barriers which basically means that such functions cannot be analytically continued past their radius of convergence. $\endgroup$
    – gone
    Jan 22, 2014 at 22:46
  • $\begingroup$ If it helps, I've also seen this condition referred to as a lacunary function. $\endgroup$
    – Matt Groff
    Jan 27, 2014 at 3:01

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