We have the known BBP(Bailey–Borwein–Plouffe)-type formulas,

$$\ln3 = \sum_{n=0}^\infty\frac{1}{2^{2n}}\left(\frac{1}{2n+1}\right)$$

$$\ln5 = \frac{1}{2^2}\sum_{n=0}^\infty\frac{1}{2^{4n}}\left(\frac{2^2}{4n+1}+\frac{2^2}{4n+2}+\frac{1}{4n+3}\right)$$

However, I noticed that if we define the function,

$$R\big(a,b\big) = \sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{2^{a-1-j}}{an+j}+\sum_{k=1}^{a/b-1}(-1)^{k+1}\frac{2^{a-1-bk}}{an+bk}\right)\tag1$$

then it seems BBP-type formulas for *Fermat numbers* $2^{2^m}+1$ with $m>0$ have a common form,

$$\ln 5 = \frac1{2^{2}}R\big(2^2,2^1\big)$$

$$\ln 17 = \frac1{2^{13}}R\big(2^4,2^2\big)$$

$$\ln 257 = \frac1{2^{252}}R\big(2^8,2^3\big)$$

$$\color{brown}{\ln 65537 \overset{?}= \frac1{2^{65531}}R\big(2^{16},2^4\big)}$$

Q: Is the formula for $p=65537$ true?

I've used Mathematica to verify the $p=5,\,17,\,257$ to hundreds of decimal digits, and also $p=65537$ using its initial terms, but how do we rigorously prove that $(1)$ is true for Fermat numbers $>3$?

**P.S.** In this paper, the authors were not(?) able to find $p=65537$ and is missing in the list.