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Is the continued fraction $\left[1;2,2,3,3,3,4,4,4,4\dots\right]$, where every positive integer $n$ is repeated $n$ times in order starting at $1$, a known value? What properties does it have? I've been unable to find any reference to this number anywhere (with an admittedly limited search).

The only observation I've been able to make is that its slightly less than $\sqrt2$, which makes sense as the continued fraction expansion of $\sqrt2$ is $[1;\bar2]$, and the above series diverges from that one quite slowly (and the elements of the series have less influnce over the final value the later into the series they are).

hakr14
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    We need a continued fractions equivalent of OEIS. – J.G. Apr 12 '21 at 06:32
  • @J.G. You could argue every integer sequence can be a continued fraction... (related OEIS sequence btw) – hakr14 Apr 12 '21 at 06:36
  • Yeah, we probably just need the OEIS to add continued fraction information. I already checked they didn't for that particular sequence. – J.G. Apr 12 '21 at 06:54
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    Numerical value is $$1.410804373305871560545114955738797442117871310429336631975\cdots$$ Wolfram Alpha's output of possible closed forms does not indicate something special about this number. Apart from its curious continued fraction representation, it seems to be just some (probably transcendental) real number. – Peter Apr 12 '21 at 07:28
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    $\frac{691+246 \pi +388 \pi ^2}{122+125 \pi +328 \pi ^2}$ is in error of $2.61\times 10^{-19}$ – Claude Leibovici Apr 12 '21 at 08:21
  • @ClaudeLeibovici How did you come up with this? – user2661923 Apr 12 '21 at 08:29
  • @user2661923 Small program of mine using a bunch of heuristics – Claude Leibovici Apr 12 '21 at 08:30
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    @hakr14 Using the lindep-function of PARI/GP , you can find such approximations as the one given by Claude : If $x$ is the number to be approximated and you use the base , lets say , $1,\pi,\pi^2$ , then we have $\frac{A+B\pi+C\pi^2}{D+E\pi+F\pi^2} \approx x$ , so $A+B\pi+C\pi^2-Dx-E\pi x-F\pi^2x \approx 0$ , so "lindep([1,Pi,Pi^2,-x,-x * Pi,-x * Pi^2],k)" gives the coefficients you need. $k$ determines the accuracy, it can as well be omitted. – Peter Apr 12 '21 at 10:40
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    @ClaudeLeibovici: Why is that approximation surprising? You have 18 digits in that expression, so we expect to be able to get ≈18 digits of accuracy.. We just need to try a few to find one. No? – user21820 Apr 12 '21 at 16:44

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