Re: 2.058833711777943011170351... = ? continued fraction convergents//
- From: raymond manzoni <raymond.manzoni@xxxxxxxxxx>
- Date: Tue, 02 Jan 2007 23:10:35 +0100
raymond manzoni a écrit :
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
So that the inspirating problem was probably the evaluation of
sum_{n>=0} 1/F_{2^n} = (7 - sqrt(5))/2
(Millin “Problem H–237”, The Fibonacci Quarterly 1974)
http://www.mathpropress.com/stan/bibliography/reciprocalGeomProgr.pdf
Note that we have too :
sum_{n=0}^m 1/F_{2^n} = 3 - F_{2^m-1}/F_{2^m}
http://pweb.nju.edu.cn/zwsun/39f.pdf
That should help to find an exact answer for the last one... if this may be done.
Fine search!
Raymond
.
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