10

(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?

Vladimir Reshetnikov
  • 32,650
  • 1
    Related papers: http://algo.inria.fr/seminars/sem92-93/allouche.pdf, https://pdfs.semanticscholar.org/a4dc/24980aad45813df581c6e65669f508b30fc4.pdf, https://arxiv.org/abs/1709.03398, https://arxiv.org/abs/1406.7407, https://arxiv.org/abs/1709.04104 – Vladimir Reshetnikov Apr 09 '18 at 01:58
  • 1
    Related questions: https://math.stackexchange.com/q/29234/19661, https://math.stackexchange.com/q/924601/19661 – Vladimir Reshetnikov Apr 11 '18 at 18:52

1 Answers1

3

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 $\tfrac 1{2\sqrt2}.$

Alex Ravsky
  • 106,166
  • 1
    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: https://arxiv.org/pdf/1709.03398.pdf – Klangen Mar 11 '19 at 09:12