Product of odd evil numbers over the product of odd odious numbers

This follows easily from corollary 2.4(b) of "More Infinite Products: Thue-Morse and the Gamma function" (arXiv:1709.03398) by Jean-Paul Allouche, Samin Riasat, and Jeffrey Shallit. The statement there is

$$ \prod_{n \ge 0} \left( {4n+1 \over 4n+3} \right)^{(-1)^{t_n}} = {1 \over 2} $$

where $t_n$ is the Thue-Morse sequence, i. e. $t_n$ is the sum of the bits of the binary expansion of $n$. Your desired result follows by taking the reciprocal of both sides to get

$$ \prod_{n \ge 0} \left( {4n+3 \over 4n+1} \right)^{(-1)^{t_n}} = 2$$

and some rewriting on the left-hand side to get a product over odd numbers.

I haven't solved this in detail, but I'm sure I know how to go about it. There are some facts about evil and odious numbers that I have found experimentally by computer. Once these are verified, they will give a complete solution.

Look at "On Infinite Products Associated with Sums of Digits," by J.O. Shallit. For $k\geq2$ a positive integer, the author defines $s_k(n)$ to be the sum of the digits of the nonnegative integer $n$ when written in base $k$. Theorem $3$ in the paper is

Let $k$ be an even positive integer and let $m_i, n_i$ be such that $2i\leq m_i,n_i<2(i+1),$ and $s_k(m_i)\equiv0\pmod2$ and $s_k(n_i)\equiv1\pmod2.$ Then $$ \prod_{i=0}^{\infty}{1+m_i\over 1+n_i}={\sqrt{k}\over k}$$

If we let $k=2,$ and let $m_i$ and $n_i$ be the evil and odious integers, respectively, this gives $$ \prod_{i=0}^{\infty}{1+m_i\over 1+n_i}={1\over \sqrt2} \tag{1} $$

I haven't proved that the growth conditions are satisfied, just verified them numerically for $0\leq i<2^{20},$ but I would expect that it isn't very hard to do this, using the Thue-Morse sequence.

Notice that if instead of running over the evil and odious numbers, $m_i$ and $n_i$ ran over the even evil and odious numbers, we would have the reciprocal the product we want. Therefore, let us divide the product in $(1)$ by $$ \prod_{i=0}^{\infty}{1+m_i'\over 1+n_i'}\tag{2}$$ where $m_i',n_i'$ run over the odd evil and odious numbers, respectively. Say $m_i'=2c_i+1, n_i'=2d_i'+1,$ so that $$ {1+m_i'\over1+n_i'}={1+c_i\over1+d_i}\tag{3}$$

The $c_i$ are precisely the odious integers, and the $d_i$ are precisely the even integers, so that the product in $(2)$ is the reciprocal of the product in $(1)$ or $1/\sqrt2,$ and $(1)$ divided $(2)$ is $1/2,$ which is just what we want. That the $c_i$ and $d_i$ are as claimed follows from the fact that $2n+1\mapsto n$ is a bijection from the odd positive integers onto to nonnegative integers, and the recurrence $t_{2n+1}=1-t_n$ for the Thue-Morse sequence.

I'll leave the growth conditions for you to verify.

I should mention that in the proof of Theorem $1$, Shallit appeals to Prouhet's solution of the Prouhet-Tarry-Escott problem, which apparently depends on the Thue-Morse sequence.