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

(related to my previous questions$^{[1]}$$\!^{[2]}) Let's define the signed Thue–Morse sequence t_n by the recurrence$$t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor},\tag1$$or by the generating function$$\sum_{n=0}^\infty t_n \, x^n=\prod_{n=0}^\infty\left(1-x^{2^n}\right).\tag{$1^\prime$}$$It seems that the following conjectures hold:$$\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+\tfrac12\right)^{t_k}\stackrel{\color{gray}?}=\frac12\tag2\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+1\right)^{t_k}\stackrel{\color{gray}?}=\frac1{\sqrt2}\tag3\lim_{n\to\infty}\prod_{k=0}^{2^n-1}\left(k+1\right)^{(-1)^k\,t_k}\stackrel{\color{gray}?}=\frac1{2\sqrt2}\tag4$\$ How can we prove these? Are there any other limits of products similar to these?

## 1 Answer

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 $$\tfrac 1{2\sqrt2}.$$

• The index splitting technique is indeed the correct way for calculating such products. As a note, I would like to mention the (fantastic) paper by Allouche, Shallit and Riasat, which derives many similar formulas: arxiv.org/pdf/1709.03398.pdf – Klangen Mar 11 at 9:12