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What is the limit of the continued fraction whose partial denominators are the composites?

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The posting says "below", but it isn't there. Let's see how it looks here: $1+\cfrac{1}{4+\cfrac{1}{6+\cfrac{1}{8+\cfrac{1}{9+ \cfrac{1}{10+\cfrac{1}{12+\cfrac{1}{14+\cfrac{1}{15+\cdots}}}}}}}}$ –  Michael Hardy Nov 18 '11 at 1:58
This question is the dual of this question. –  Guess who it is. Nov 18 '11 at 2:33
Maybe I should have omitted "$1+{}$" at the beginning. –  Michael Hardy Nov 18 '11 at 2:55

2 Answers 2

You mean $1/(1+1/(4+1/(6+1/(8+1/(9+1/\ldots)$? It is approximately


but I doubt that there is a "closed form".

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Actually your numerical quote is for the reciprocal of the continued fraction. –  Sasha Nov 17 '11 at 21:01
The Plouffe inverter recognizes neither this number nor its reciprocal. –  Gerry Myerson Nov 17 '11 at 22:55

That the continued fraction converges can be easily established through the Śleszyński-Pringsheim theorem.

For reference, here's some Mathematica code for computing this number to prec significant figures, which is similar to the code given here:

Composite[n_Integer?Positive] := FixedPoint[(n + PrimePi[#] + 1) &, n]

prec = 500;
y = N[1, prec + 5]; c = y; d = 0; k = 1;
While[True, p = Composite[k];
  c = p + 1/c; d = 1/(p + d);
  h = c*d; y *= h;
  If[Abs[h - 1] <= 10^(-prec), Break[]];
N[y, prec]

which yields


Its reciprocal is


Gerry has said that Plouffe's inverter can't recognize these numbers; neither can the ISC.

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I didn't say Plouffe's inverter doesn't work; I just said it doesn't recognize either of these numbers. –  Gerry Myerson Nov 18 '11 at 3:38
I mean it doesn't work on these numbers. But sure, I'll clarify... –  Guess who it is. Nov 18 '11 at 3:45

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