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

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

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

If we let $b_k = -2\cos(2k\,\pi/5)$, I noticed this can be expressed using more familiar integers,

$$u_2 = 5^2\ln\left(\frac{(5b_1+5^2+1)^{b_2}\,(5b_2+5^2+1)^{b_1}}{4^2}\right)= 2.833601\dots$$

and 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 the radicals and integers can be de-mystified as $(57+5\sqrt{5})(57-5\sqrt{5}) = 4(781) = 5^5-1$. 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 $d_{13} = 13^2+1 = 170$.

Solution 1:

This is a late answer, but I hope it helps. In page 20 of the link, there's a formula $$\sum_{n=0}^\infty\frac1{b^n(dn+c)}=-\frac1db^{c/d}\sum_{a=0}^{d-1}e^{-2\pi iac/d}\log\left(1-\frac{e^{2\pi ia/d}}{b^{1/d}}\right)$$which can be proved by the Taylor expansion of $-\log(1-x)$. From this we have \begin{align}v_k&=\sum_{n=0}^\infty\frac{1}{p^{pn}}\left(\frac{p^{p-2k}}{pn+k}+\frac{1}{pn+p-k}\right)\\&=-p^{p-k-1}\sum_{a=0}^{p-1}(e^{-2\pi iak/p}+e^{2\pi iak/p})\log\left(1-\frac{e^{e\pi ia/p}}{p}\right)\\&=-p^{p-k-1}\sum_{a=0}^{p-1}\cos\frac{2\pi ka}{p}\log\left(p^2+1-2p\cos\frac{2\pi a}{p}\right)\\&=-p^{p-k-1}\left(\sum_{a=1}^{(p-1)/2}2\cos\frac{2\pi ka}{p}\log\left(p^2+1-2p\cos\frac{2\pi a}{p}\right)+\log(p-1)^2\right)\end{align} Let $c_k=-2\cos(2k\pi/p)$, $d=p^2+1$. Then the above can be expressed as $$v_k=p^{p-k-1}\log\left(\frac{1}{(p-1)^2}\prod_{a=1}^{(p-1)/2}(pc_a+d)^{c_{ka}}\right)$$ So the subscripts are all $ka$. Let $p=11$. Then $$v_1=1,2,3,4,5\\v_2=2,4,6,8,10\\v_3=3,6,9,12,15\\v_4=4,8,12,16,20\\v_5=5,10,15,20,25$$ Notice that $c_k=c_{p-k}=c_{p+k}$. The above can be transformed as $$v_1=1,2,3,4,5\\v_2=2,4,5,3,1\\v_3=3,5,2,1,4\\v_4=4,3,1,5,2\\v_5=5,1,4,2,3$$The fact that they form a Latin square follows from the fact that the $ka$ s are all different by modulo $p$.