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.


Yes, this actually holds for all Fermat numbers! Let's start with the identity $$\log 2=\sum_{k=1}^{\infty}\frac{1}{k2^k}$$ and try to work out an expression for $$\log (2^{2^s}+1)=2^{s}\log 2+\log\left(1-\frac{1}{2^{2^{s+1}}}\right)-\log\left(1-\frac{1}{2^{2^s}}\right)$$ $$=2^s\left(\sum_{k=1}^{\infty}\frac{1}{k2^k}\right)-\left(\sum_{k=1}^{\infty}\frac{1}{k2^{k2^{s+1}}}\right)+\left(\sum_{k=1}^{\infty}\frac{1}{k2^{k2^{s}}}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}-\sum_{h=1}^{2^{2^s-s-1}}\frac{2^{-h2^{s+1}}}{n2^{2^{s}-s-1}+h}+\sum_{l=1}^{2^{2^s-s}}\frac{2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}-\sum_{h=1}^{2^{2^s-s-1}}\frac{2^{-(2h)2^{s+1}}}{n2^{2^{s}-s}+2h}+\sum_{l=1}^{2^{2^s-s}}\frac{2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}+\sum_{l=1}^{2^{2^s-s}}\frac{(-1)^{l+1}2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ here we see that the terms at $j=2^{2^s}$ and $l=2^{2^s-s}$ cancel out, so after factoring out $\frac{1}{2^{2^{2^s}-s-1}}$ we are left with the defining expression of $R$ with $a=2^{2^s}, b=2^s$: $$\log(2^{2^s}+1)=\frac{1}{2^{2^{2^s}-s-1}}R(2^{2^s},2^s).$$

  • 1
    $\begingroup$ Beautiful! In his compendium, Bailey (the first B in BBP), listed $p=3,5,7,17,257$ but apparently forgot $p=65537$ since he included larger primes in his list. $\endgroup$ – Tito Piezas III Jul 5 '17 at 13:52
  • $\begingroup$ Can you kindly look at this post as the counterpart formula for Mersenne numbers? Maybe a small tweak to your proof can settle that as well. $\endgroup$ – Tito Piezas III Jul 7 '17 at 16:32

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