Define three generalized hypergeometric functions (which differs in the starting numerator in blue): $$A_k =\,_kF_{k-1}\left(\left.\begin{array}{c} 1,\frac{\color{blue}{k-2}}{k-1},\frac{k-1}{k-1},\dots\\ \frac{k+1}{k}, \frac{k+2}{k},\dots \end{array}\right| \frac{(k-1)^{k-1}}{k^k}\right)$$

$$B_k =\,_kF_{k-1}\left(\left.\begin{array}{c} 1,\frac{\color{blue}{k-1}}{k-1},\frac{k}{k-1},\dots\\ \frac{k+1}{k}, \frac{k+2}{k},\dots \end{array}\right| \frac{(k-1)^{k-1}}{k^k}\right)$$

$$C_k =\,_kF_{k-1}\left(\left.\begin{array}{c} 1,\frac{\color{blue}{k}}{k-1},\frac{k+1}{k-1},\dots\\ \frac{k+1}{k}, \frac{k+2}{k},\dots \end{array}\right| \frac{(k-1)^{k-1}}{k^k}\right)$$

The smallest common non-trivial case would then be $k=3$, $$A_3={_3F_2}\left(1,\tfrac12,\tfrac22;\ \tfrac43,\tfrac53;\ \tfrac4{27}\right)$$ $$B_3={_3F_2}\left(1,\tfrac22,\tfrac32;\ \tfrac43,\tfrac53;\ \tfrac4{27}\right)$$ $$C_3={_3F_2}\left(1,\tfrac32,\tfrac42;\ \tfrac43,\tfrac53;\ \tfrac4{27}\right)$$

The first two were addressed in this and this posts by different OPs from different perspectives, but a common approach may be possible. First, let $x_n$ be the roots of $x^k-x+1=0$.

Question 1:

Is it true, $$\frac{A_k}{(k-1)(k-2)} = \int_1^\infty \frac{-k+(k-1)x}{x^k-x+1}dx =-\sum_{n=1}^k x_n\ln(1-x_n)\tag1$$ Note: This implies a simple solution to the case $k=3$ discussed in by Reshetnikov in this post as, $$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}= -x_1\ln(1-x_1)-x_2\ln(1-x_2)-x_3\ln(1-x_3) = 0.5179778\dots$$ where the $x_n$ are the three roots of $x^3-x+1=0\;^\color{red}*$ and which is the minpoly of the negated plastic constant.

Question 2:

Is it true, $$\frac{B_k}{k} = \int_1^\infty\frac1{x(x^k-x+1)}dx=-\sum_{n=1}^k \frac{\ln(1-x_n)}{-k+(k-1)x_n}\tag2$$ Note: This implies a similarly simple solution to the case $k=3$ in this post,

$$\sum_{n=1}^\infty\frac1{n\binom{3n}n} = \frac{\ln(1-x_1)}{3-2x_1}+ \frac{\ln(1-x_2)}{3-2x_2}+ \frac{\ln(1-x_3)}{3-2x_3}=0.371216\dots$$ and the $x_n$ again are the three roots of $^\color{red}*$.

Question 3:

See this post.


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