TITLE: An asymptotic series for $\small\prod_{k=n}^\infty\operatorname{sinc}\left({2^{-k}}\pi\right),\,n\to\infty$
QUESTION [12 upvotes]: Using empirical methods, I conjectured that$^{[1]}$$\!^{[2]}$
$$\small\prod_{k=n}^\infty\operatorname{sinc}\left({2^{-k}}\pi\right)=1-\frac{2\pi^2}9\,4^{-n}+\frac{38 \,\pi ^4}{2025}\,4^{-2n}-\frac{2332\,\pi ^6}{2679 075}\,4^{-3 n}+\frac{265618\,\pi^8
   }{10247461875}\,4^{-4 n}+O\!\left(4^{-5 n}\right)\!.$$
Can we prove this? Can we find more coefficients in this expansion, or a general formula for those coefficients?

REPLY [7 votes]: Writing $\operatorname{sinc} x$ as an infinite product, taking the logarithm and changing the summation order in the triple sum gives
$$\ln \prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) =
\sum_{k \geq n} \ln \prod_{l \geq 1} \left( 1 - \frac {2^{-2 k}} {l^2} \right) =
-\sum_{k \geq n} \sum_{l \geq 1} \sum_{m \geq 1}
 \frac 1 m \left( \frac {2^{-2 k}} {l^2} \right)^{\!m} = \\
\sum_{m \geq 1} c_m \frac {2^{-2 m n}} {m!},
\quad c_m = -\frac {2^{2 m} \Gamma(m) \zeta(2 m)} {2^{2 m} - 1}.$$
Then, by Faa di Bruno's formula,
$$\prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) =
1 + \sum_{m \geq 1} \sum_{1 \leq l \leq m}
 B_{m, l}(c_1, \ldots, c_{m - l + 1}) \frac {2^{-2 m n}} {m!} = \\
1 + \sum_{m \geq 1} B_m(c_1, c_2, \ldots, c_m) \frac {2^{-2 m n}} {m!},$$
where $B_{m, l}$ and $B_m$ are the Bell polynomials.