Re: 2.058833711777943011170351... = ? continued fraction convergents//

Robert Israel a écrit :
In article <4vv0i1F1dhur0U1@xxxxxxxxxxxxxxxxxx>,
The Last Danish Pastry <clivet@xxxxxxxxx> wrote:
"Proginoskes" <CCHeckman@xxxxxxxxx> wrote in message news:1167458262.298359.10060@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

don.lotto@xxxxxxxxxxxxxxx wrote:
don.lotto@xxxxxxxxxxxxxxx wrote:
N. Silver wrote:
Proginoskes wrote:

2 + (1 / (17 - (1 / (340 - (1 / (8.5 + (1 / 51))))))) = 2.05883371

fermat gauss numbers on casio fx 82 schools calc.
from memory. signs may be wrong.
Exact value: 2.058833712. No good.

regular n-gon.
multiples 2 3 5 17
= 2^(2^n) +1.

Re: 2.058833711777943011170351... = ? continued fraction convergents//
I tried that, but I guess I didn't mention it. The continued fraction
starts out as
[2, 16, 1, 338, 1, 7, 2, 1964179, 1, 1, 8, 169, 1, 3, 1, 6,
22000684073033, ...]
So, it appears that you have now revealed that your
2.058833711777943011170351...
is actually
2.0588337117779430111703514900024026360848819899999645686770...

... and that it is the sum of a series

1 + 1/1 + 1/17 + 1/(17*53*109) + 1/(17*53*109*2269*4373*19441) + ...

But I can't guess what the next term will be.


Nice! The next one is :
+1/(17*53*109*2269*4373*19441*448607550257*16000411124306403070561) of course!

the denominators are the products of a(n) with
a(n)= a(n-1)^3 + 3a(n-1)^2 - 3
(well 1/1 should be 1/2 but....)

the sum is simply :
1 + 2* sum_{n>=0) 1/Fib(3^n)

http://www.research.att.com/~njas/sequences/A002814

Raymond


Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
.

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