The number is the infinite product:$$\frac{2 \cdot 4\cdot 8 \cdot 16 \cdots}{1\cdot 3\cdot 7\cdot15 \cdots},$$ or about $3.4628.$
I have only proven this number is finite. Can you either prove it is irrational or express it as a fraction with integer numerator and denominator?
This is equivalent to whether $$\phi\left(\frac12\right)=\prod_{n=1}^\infty\left(1-\frac1{2^n}\right)$$ is rational, where $\phi$ is Euler’s function.
But Euler’s function power series formula gives:
$$\phi\left(\frac12\right)=\sum_{n=-\infty}^{\infty}(-1)^n\frac{1}{2^{(3n^2-n)/2}}$$
I think a certain amount of fiddling should show that this value does not have a repeating base $2$ expansion, so the value is not rational.
Basically, grouping $n=2k,2k+1,-2k,-2k-1$ for $k>0$ gives $$\phi(1/2)=q+\sum_{k=1}^{\infty}\left(\frac{1}{2^{6k^2-k}}+\frac{1}{2^{6k^2+k}}-\frac{1}{2^{6k^2+5k+1}}- \frac1{2^{6k^2+7k+2}}\right)\tag1$$ where $q,$ the sum for $n=-1,0,1,$ is rational.
Now, for $m_2>m_1,$ $$\frac{1}{2^{m_1}}-\frac1{2^{m_2}}=\sum_{j=m_1+1}^{m_2}\frac1{2^j}$$
Then: $$\frac{1}{2^{6k^2+k}}-\frac{1}{2^{6k^2+5k+1}}- \frac1{2^{6k^2+7k+2}}\\=\sum_{i=6k^2+k+1}^{6k^2+5k+1}\frac{1}{2^i}-\frac{1}{2^{6k^2+7k+2}}\\=\sum_{i=6k^2+k+1}^{6k^2+5k}\frac{1}{2^{i}}+\frac1{2^{6k^2+5k+1}}-\frac{1}{2^{6k^2+7k+2}}\\=\sum_{i=6k^2+k+1}^{6k^2+5k}\frac1{2^{i}}+\sum_{j=6k^2+5k+2}^{6k^2+7k+2}\frac1{2^j}$$
So the summand in $(1)$ is $$\frac{1}{2^{6k^2-k}}+ \sum_{i=6k^2+k+1}^{6k^2+5k}\frac1{2^{i}}+\sum_{j=6k^2+5k+2}^{6k^2+7k+2}\frac1{2^j}$$
Since $6(k+1)^2-(k+1)=6k^2+11k+5>(6k^2+7k+2)+1,$ these ranges of bits do not overlap, nor are even adjacent.
In particular, the locations of the $1$ bits surrounded by $0$ bits are at only the positions $6k^2-k,$ and thus the binary expansion does not repeat.
You can easily use the same proof to show that $\phi(1/b)$ is irrational for any integer with $b>1,$ using base $b$ arithmetic.
You can give a similar argument for $b<-1,$ but need a different grouping. Basically you need to group the eight terms for $k,k+1,$ when $k$ is even.