arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 0709.4031math.NT
收藏

具有强$B$ -乘指数的无穷乘积

Infinite products with strongly $B$-multiplicative exponents

Jean-Paul Allouche, Jonathan Sondow

摘要:

令$N_{1,B}(n)$表示整数$n$的$B$进制展开式中的1的数目。Woods引入了无穷乘积$P :=\prod_{n \geq 0} (\frac{2n+1}{2n+2})^{(-1)^{N_{1,2}(n)}}$,Robbins证明了$P = 1/\sqrt{2}$。几位作者研究了相关产品。我们证明了一个技巧,证明$P^2 = 1/2$(知道$P$收敛)可以扩展到评估新产品(广义)强$B$ -乘法指数。一个简单的例子是$$ \prod_{n \geq 0} (\frac{Bn+1}{Bn+2})^{(-1)^{N_{1,B}(n)}} = \frac{1}{\sqrt B}. $$

英文摘要:

Let $N_{1,B}(n)$ denote the number of ones in the $B$-ary expansion of an integer $n$. Woods introduced the infinite product $P :=\prod_{n \geq 0} (\frac{2n+1}{2n+2})^{(-1)^{N_{1,2}(n)}}$ and Robbins proved that $P = 1/\sqrt{2}$. Related products were studied by several authors. We show that a trick for proving that $P^2 = 1/2$ (knowing that $P$ converges) can be extended to evaluating new products with (generalized) strongly $B$-multiplicative exponents. A simple example is $$ \prod_{n \geq 0} (\frac{Bn+1}{Bn+2})^{(-1)^{N_{1,B}(n)}} = \frac{1}{\sqrt B}. $$

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑