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Notational convention: $$\bigoplus_{k=0}^{\infty}a_k=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots}}}}$$

Let $$ P:=\bigoplus_{k=1}^{\infty}p_k$$

where $p_k$ is the k-th prime number.

Conjecture: $$P=e?$$

Numerical evidence: First terms:

  2+1/(3+1/(5+1/(7+1/(11+1/(13+1/(17+1/(19+1/(23+1/(29+1/(31+1/(37+1/(41)))))))))))
=2.313036736433582906383951602640999732416979599998686694360781...

Maybe after some time it will converge to 2.718...

What is known about this constant?

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    $\begingroup$ The value of this continued fraction is somewhere between $37/16\approx2.3125$ (what you get by truncating at $a_3=5$) and $266/115\approx2.31304$ (what you get by truncating at $a_4=7$. Doesn't look like it will be $e$. Why did you think it would? $\endgroup$ – Jyrki Lahtonen Sep 19 '14 at 21:11
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    $\begingroup$ The continued fraction of $e$ is known, and it's not the primes. (Apparently, $\tanh(1)=\dfrac{e^2-1}{e^2+1}$'s continued fraction is the odd integers - and I'm not entirely sure why this is - but I've never heard anything about the primes.) $\endgroup$ – Akiva Weinberger Sep 19 '14 at 21:13
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    $\begingroup$ @JeppeStigNielsen That's because they started with 0. OP is referring to the reciprocal. $\endgroup$ – Akiva Weinberger Sep 19 '14 at 21:16
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    $\begingroup$ See OEIS A06442. $\endgroup$ – Nick Matteo Sep 19 '14 at 21:17
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    $\begingroup$ This number appears here oeis.org/wiki/Continued_fractions $\endgroup$ – MPW Sep 19 '14 at 21:18
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Continued fractions don't work the way you think they do. Let $$x = a_0 + \cfrac1{a_1+\cfrac 1{a_2 +\cdots}},$$ which we will abbreviate as $x = [a_0; a_1, a_2, \ldots]$. Then probably the most important thing to know about continued fractions is $$a_0 < [a_0; a_1, a_2] < [a_0; a_1, a_2, a_3, a_4] < \ldots < x < \ldots < [a_0; a_1, a_2, a_3] < [a_0; a_1] $$

Your fraction, $P =[2; 3, 5, 7, 11, \ldots]$ has the property that

$$\frac {37}{16} = [2;3,5] < P < [2;3] = \frac 73$$ so we know that $2.3125 < P < 2.333\ldots$ and it is therefore impossible that $P=e$.

Continued fractions converge really fast. If you compute the first 13 terms and find that the result is approximately 2.31303673, then the result of carrying the fraction to infinity will be some number extremely close to 2.31303673. It is impossible that the remaining terms could budge the result as far as $2.7$. There is a theorem that if $x = [a_0; a_1, \ldots]$ and $[a_0; a_1\ldots a_k] = \frac{p_k}{q_k}$, then $$\left\lvert x - \frac{p_k}{q_k}\right\rvert < \frac1{q_kq_{k+1}} < \frac1{q_k^2}.$$ The $q_k$ increase exponentially quickly in relation to $k$, at a rate of at least $\phi^n$, where $\phi = \frac12(1+\sqrt 5)$, so the difference between $x$ and its approximation by a finite continued fraction of length $n$ is at least as fast as $O(\phi^{-2n})$. The most slowly-converging continued fraction is $[1;1,1,1,1\ldots]$, and you can see for yourself how quickly it converges to its limit, which also happens to be $\phi$.

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