# On Bailey and Crandall's sum for $\sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5}{5n+2}+\frac{1}{5n+3}\right)$

On page 20 of "On the Random Character of Fundamental Constant Expansions", Bailey and Crandall gave the rather unusual sum,

$$u_2 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5}{5n+2}+\frac{1}{5n+3}\right) = \frac{5^2}{2}\ln\left(\frac{781}{256}\left(\frac{57-5\sqrt{5}}{57+5\sqrt{5}} \right)^\sqrt{5} \right) = 2.833601\dots$$

I found its counterpart,

$$u_1 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5^3}{5n+1}+\frac{1}{5n+4}\right) = \frac{5^3}{2}\ln\left(\frac{781}{256}\left(\frac{57+5\sqrt{5}}{57-5\sqrt{5}} \right)^\sqrt{5} \right) = 125.256703\dots$$

Note that $(57+5\sqrt{5})(57-5\sqrt{5}) = 4(781) = 5^5-1$. This can be generalized though. The case $p=7$ is straightforward, but $p=11$ is a bit tricky. Let,

\begin{aligned} v_1 &= \sum_{n=0}^\infty \frac{1}{11^{11n}}\left(\frac{11^9}{11n+1}+\frac{1}{11n+10}\right)\\ v_2 &= \sum_{n=0}^\infty \frac{1}{11^{11n}}\left(\frac{11^7}{11n+2}+\frac{1}{11n+9}\right)\\ v_3 &= \sum_{n=0}^\infty \frac{1}{11^{11n}}\left(\frac{11^5}{11n+3}+\frac{1}{11n+8}\right)\\ v_4 &= \sum_{n=0}^\infty \frac{1}{11^{11n}}\left(\frac{11^3}{11n+4}+\frac{1}{11n+7}\right)\\ v_5 &= \sum_{n=0}^\infty \frac{1}{11^{11n}}\left(\frac{11}{11n+5}+\frac{1}{11n+6}\right) \end{aligned}

and define,

$$c_k = -2\cos(2k\,\pi/11)$$

$$d = 11^2+1 = 122$$

then,

\begin{aligned} v_1 &= 11^9 \ln\Big((11c_1+d)^{c_1}\,(11c_2+d)^{c_2}\,(11c_3+d)^{c_3}\,(11c_4+d)^{c_4}\,(11c_5+d)^{c_5}\,/10^2 \Big)\\ v_2 &= 11^8 \ln\Big((11c_1+d)^{c_2}\,(11c_2+d)^{c_4}\,(11c_3+d)^{c_5}\,(11c_4+d)^{c_3}\,(11c_5+d)^{c_1}\,/10^2 \Big) \\ v_3 &= 11^7 \ln\Big((11c_1+d)^{c_3}\,(11c_2+d)^{c_5}\,(11c_3+d)^{c_2}\,(11c_4+d)^{c_1}\,(11c_5+d)^{c_4}\,/10^2 \Big) \\ v_4 &= 11^6 \ln\Big((11c_1+d)^{c_4}\,(11c_2+d)^{c_3}\,(11c_3+d)^{c_1}\,(11c_4+d)^{c_5}\,(11c_5+d)^{c_2}\,/10^2 \Big) \\ v_5 &= 11^5 \ln\Big((11c_1+d)^{c_5}\,(11c_2+d)^{c_1}\,(11c_3+d)^{c_4}\,(11c_4+d)^{c_2}\,(11c_5+d)^{c_3}\,/10^2 \Big) \\ \end{aligned}

For clarity, the exponent $c_k$ have subscripts,

\begin{aligned} u_1 &= 1,2,3,4,5\\ u_2 &= 2,4,5,3,1\\ u_3 &= 3,5,2,1,4\\ u_4 &= 4,3,1,5,2\\ u_5 &= 5,1,4,2,3\\ \end{aligned}

Question: What is the formula for the sequence of exponent subscripts? (I noticed it is a Latin square, but how do you generate it?)

P.S. The case $p=13$ is analogous, and uses $d_{13} = 13^2+1 = 170$.

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I was going to email Richard Crandall (the co-author of the paper cited) but found out from wikipedia that he died Dec 2012. :( –  Tito Piezas III Sep 2 '13 at 1:40