Closed form for $\prod_{n=0}^\infty (1-z^{2^n})$

Question

Is there a closed form for the product $$f(z) = \prod_{n=0}^\infty 1-z^{2^n}$$ either as a formal power series or as an analytic function in the disk $|z| < 1$? It's not hard to see that Taylor series coefficients of $f$ about 0 are all $\pm 1$: $$f(z) = 1 - z - z^2 + z^3 - z^4 + z^5 + z^6 - z^7 + \dotsb$$ and that they form a pattern in blocks, so to speak: $$+- \quad\to\quad +--+ \quad\to\quad +--+-++- \quad\to\quad \dotsb$$ But I don't know much else. I came across this infinite product when I incorrectly transcribed an exercise from a book, wasting a good deal of time before returning to the book to ask it, "are you sure?". (The exercise was to prove that $\prod_{n=0}^\infty 1 + z^{2^n} = (1-z)^{-1}$, which is easy.)

Answer 1

Converting my comment from 2016 into an answer, with elaborations:

The coefficients of $z^n$ in $f(z)$ is $-1$ to the power of the number of $1$s in the binary expansion of $n$; this is a slight variant of the Thue-Morse sequence with $1$ and $-1$ instead of $0$ and $1$ (A106400 on the OEIS). The coefficient is $1$ if $n$ is "evil" (has an even number of $1$s; A001969 on the OEIS) and $-1$ if $n$ is "odious" (has an odd number of $1$s; A000069 on the OEIS). The reciprocal

$$\frac{1}{f(z)} = \prod_{n \ge 0} \frac{1}{1 - z^{2^{n}}}$$

is the generating function for the sequence $c_n$ counting the number of partitions of $n$ into powers of $2$, which is A018819 on the OEIS. $c_{2n}$ is A000123 on the OEIS because $c_{2n+1} = c_{2n}$.

According to Kachi and Tzermias' On the $m$-ary partition numbers this problem was first studied by Euler. Mahler showed in On a special functional equation that

$$c_n = e^{O(1)} 2^{ -{k \choose 2}} \frac{n^k}{k!}$$

where $k$ is the unique positive integer satisfying $2^{k-1} k \le n < 2^k (k+1)$. This gives $k \approx \frac{W(n \log 2)}{\log 2}$ where $W(x) \sim \log x$ is the Lambert W function, which lets us write down an asymptotic for the logarithm of $c_n$ the leading term of which is

$$\log c_n \sim \frac{(\log n)^2}{2 \log 2}.$$

de Brujin's On Mahler's partition problem analyzes the asymptotics more precisely.

These asymptotics imply that $c_n$ grows faster than polynomially but slower than exponentially and so rule out $f(z)$ having a particularly simple closed form; for example it follows that $f(z)$ cannot be rational (although this has an easier proof) or more generally meromorphic or algebraic.

Answer 2

You can find a sort of a closed form, by the standard trick of taking logs. Note that $$\log(1-z) = -\sum_{k=0}^\infty \frac{z^k}k.$$ If you take the log of your product, the coefficient of $z^k$ will be equal to $$-\sum_{m=1}^{2^{\lceil \log_2 k\rceil}} \frac{1}{m}=-H_{2^{\lceil \log_2 k\rceil}},$$ where $H_n$ are the harmonic numbers.

Answer 3

Notice that if $$f(z)=\prod_{n\ge0}(1-z^{2^n}),$$ then $$f(z)=(1-z)f(z^2).\tag1$$ Then define the sequence $a_n$ to satisfy $$f(z)=\sum_{n\ge0}a_nz^n.$$ Then from $(1)$ we have $$\sum_{n\ge0}a_nz^n=(1-z)\sum_{n\ge0}a_nz^{2n}.$$ This is equivalent to $$\sum_{n\ge0}a_nz^n=b_0+\sum_{n\ge1}(b_n-b_{n-1})z^n,$$ where $b_n=[2|n]\cdot a_{n/2}$, with $[2|n]=\frac{1+(-1)^n}{2}$ being the Iverson bracket. Since $f(0)=1$, we have $a_0=b_0=1$, and for $n\ge1$, $$a_n=a_{n/2}[2|n]-a_{(n-1)/2}[2|n-1].\tag2$$ For $n=2k$, $$a_{2k}=a_k[2|2k]-a_{k-1/2}[2|2k-1]=a_k.$$ Similarly, for $n=2k+1$, $$a_{2k+1}=a_{k+1/2}[2|2k+1]-a_{k}[2|2k]=-a_k.$$ As you saw, it is then easy to see that $|a_k|=1$. It then follows directly that $a_{2^n}=-1$.

We can find some special cases of $a_m$ where $m=2^k$, such as $$\begin{align} a_{2m+1}&=1\\ a_{4m+2}&=1\\ a_{4m+3}&=-1\\ a_{8m+6}&=-1\\ a_{8m+7}&=1\\ a_{16m+14}&=1\\ a_{16m+15}&=-1\\ a_{32m+30}&=-1\\ a_{32m+31}&=1\\ &... \end{align}$$ which all follow from $a_m=-1$.

Clearly, $a_n$ expresses something about the ways in which one can write $n$ as $a2^k+b$, for very specific $a,b$, but other than that, it is proving exceedingly difficult to find a general closed form for $a_n$. I doubt a closed form for $f(z)$ exists.