What is the limit of the continued fraction whose partial denominators are the composites?
 71,951
 6
 191
 335
 51
 1

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 01:58

[This question](http://math.stackexchange.com/questions/63865) is the dual of this question. – J. M. ain't a mathematician Nov 18 '11 at 02:33

Maybe I should have omitted "$1+{}$" at the beginning. – Michael Hardy Nov 18 '11 at 02:55
2 Answers
You mean $1/(1+1/(4+1/(6+1/(8+1/(9+1/\ldots)$? It is approximately
1.240193472713540402907659901653619438804869904021954760571414246112567530857899396707443536897205404
but I doubt that there is a "closed form".
 416,382
 24
 302
 603

Actually your numerical quote is for the reciprocal of the continued fraction. – Sasha Nov 17 '11 at 21:01

1The Plouffe inverter http://pi.lacim.uqam.ca/eng/ 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 ŚleszyńskiPringsheim 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[]];
k++];
N[y, prec]
which yields
1.2401934727135404029076599016536194388048699040219547605714142461125675308578993967074435368972054040499444569870631879216700131159033931437886878647547511888183754244113415120681779806667661819237665606323444679166964461726592199240306129802843968806256258252294199900746887849052219088971773650043418994122502389225153031976851620685269261818925150482368821451503313305347497194089489582641069356680422299551667227442255378383831719803128578492721019246089346156505612206821587675561358847518295030
Its reciprocal is
0.80632580480527957525355985703886211400308652751886019336591742291710688332574070079436098605441519195401666000252471896335967661438294029308273049422789739675889321789197956373719153024896914087422889469613361550097143208723090059691265863842011401757403471316704470493263754267277999620082114009184258161800044132073167763756696153308835343268670549110722478950038217797937443267139158452067897088395434297330559287749430628860526069885283389279384016537671968819880420551471111687607529969189158471
Gerry has said that Plouffe's inverter can't recognize these numbers; neither can the ISC.
 71,951
 6
 191
 335

1I 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 03:38

I mean it doesn't work on these numbers. But sure, I'll clarify... – J. M. ain't a mathematician Nov 18 '11 at 03:45