A few conjectured limits of products involving the Thue–Morse sequence

2) For $n\ge 1$ we have

$$\prod_{k=0}^{2^n-1}\left(k+\tfrac12\right)^{t_k}=$$ $$\prod_{k=0}^{2^{n-1}-1}\left(2k+\tfrac12\right)^{t_{2k}}\left(2k+1+\tfrac12\right)^{t_{2k+1}}=$$ $$\prod_{k=0}^{2^{n-1}-1}\left(2k+\tfrac12\right)^{t_k}\left(2k+1+\tfrac12\right)^{-t_k}=$$ $$\prod_{k=0}^{2^{n-1}-1}\left(\frac{2k+\tfrac12}{2k+1+\tfrac12}\right)^{t_k}=$$ $$\prod_{k=0}^{2^{n-1}-1}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}.$$

So, according to the introduction and Lemma 1 from the last paper you referenced (“Infinite products involving binary digit sums” by Samin Riasat), the left hand side of (2) equals

$$\prod_{k=0}^{\infty}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}=$$ $$\frac13\prod_{k=1}^{\infty}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}=$$ $$\frac13f\left(\frac14,\frac34\right)=\frac13\cdot\frac32=\frac12.$$

3) Similarly to the previous case we can show that the left hand side of (3) equals $\tfrac12 f\left(\tfrac 12,1\right)=\tfrac 1{\sqrt2}$.

1) Here preliminary calculations are a bit longer. For $n\ge 2$ we have

$$\prod_{k=0}^{2^n-1}(k+1)^{(-1)^kt_k}=$$ $$\prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_{2k}}(2k+1+1)^{-t_{2k+1}}=$$ $$\prod_{k=0}^{2^{n-1}-1}(2k+1)^{t_k}(2k+2)^{t_k}=$$ $$\prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_{2k}}((4k+3)(4k+4))^{t_{2k+1}}=$$ $$\prod_{k=0}^{2^{n-2}-1}((4k+1)(4k+2))^{t_k}((4k+3)(4k+4))^{-t_{k}}=$$ $$\prod_{k=0}^{2^{n-2}-1}\left(\frac{k+\tfrac14}{k+\tfrac34}\right)^{t_k}\left(\frac{k+\tfrac12}{k+1 }\right)^{t_k}.$$

Thus the left hand side of (1) equals the product of the left hand sides of (2) and (3), which is $\tfrac 1{2\sqrt2}.$