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Fibonacci Fractions (Posted on 2005-03-09) Difficulty: 3 of 5
What is the sum of 0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+ 0.000008+ ..., where each term is the n-th Fibonacci number, shifted n places to the right (that is, divided by 10^n)?

See The Solution Submitted by e.g.    
Rating: 3.0000 (2 votes)

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Solution re: Extended precision using UBASIC -- base 12 error in added matter | Comment 7 of 17 |
(In reply to Extended precision using UBASIC by Charlie)

In base-12, the actual repetition is

0916030534351145038167938931297709923664122137404580152671755725190839694656488
549618320610687022900763358778625954198473282442748 / 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999

=
 12/131

This is 12/(12^2-12-1); in base 10 we had 10/(10^2-10-1), so that might be the pattern.

The corrected base-12 program is

   10   F1=1:F2=1:Pwr=1/144
   15   Tot=1/12+1/144
   20   while Pwr>0
   30    F3=F1+F2
   35     PPwr=Pwr
   40    Pwr=Pwr/12
   50    Tot=Tot+F3*Pwr
   60    F1=F2:F2=F3
   70    print Tot
   80   wend
   90   print F1,int(-log(PPwr)/log(10)),int(-log(F1)/log(10)-log(PPwr)/log(10))

the corrected line being line 15

 


  Posted by Charlie on 2005-03-09 20:17:56
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