20030326, 12:52  #1 
Mar 2003
3^{4} Posts 
Continued Fraction of Primes
If one knows
x=1/(2 + 1/(3 + 1/(5 +1/(7 +1/(11 ..., he can easily produce the list of all primes, but what does x equal to by itself? 
20030326, 13:37  #2 
Dec 2002
Frederick County, MD
2·5·37 Posts 
0.4323
3208718590 2868909253 7932419999 6370511089 6877651310 3281520671 5855390511 5295886642 4773023467 5307312901 3588747517 1102192547 3474173059 9816815325 2537010284 6860319246 0457044667 2860224884 0679362020 1938436437 9879295524 6786129609 7638935269 4027752231 9731978458 6355957940 3620206633 8633654544 8951089096 5971586278 7332585763 6862001836 7995212808 7865043794 6101266432 6042252640 0822552675 2215113354 1703783531 9471839... Works Cited "Prime Curios!" http://primes.utm.edu/curios/page.php?number_id=2750 
20030327, 13:41  #3 
Mar 2003
3^{4} Posts 
That is fake!
Is it possible to calculate the fraction without the beforehand knowing of all primes ? 
20030327, 14:30  #4  
Dec 2002
Frederick County, MD
562_{8} Posts 
Quote:


20030327, 15:58  #5  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
Have you ever studied infinite series in a math class? There are techniques for calculating the sum of an infinite series once the general formula for each term is known, even though not all the specific values are known. Continued fractions can be similarly calculated. 

20030329, 07:16  #6  
Mar 2003
81_{10} Posts 
Quote:
None can give a form for nmember of such kinda series, which convergence to continued fraction of all primes. 

20030329, 13:13  #7 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Okay, you're right. It's not possible. The number is fake. I was blowing smoke, and you called me on it.

20030329, 19:27  #8 
Aug 2002
2×101 Posts 
I guess he can take it up with Michael Hartley now.

20030330, 02:19  #9  
Dec 2002
Frederick County, MD
2×5×37 Posts 
Quote:


20030413, 14:56  #10 
Sep 2002
Vienna, Austria
11011011_{2} Posts 
Consider the continued fration
x=1+1/(2+1/(3+1/(4+1/(......... I've calculated the value of x and got this result: [code:1]x=1.4331274267 2231175831 7183455775 9918204315 1276790598 0523434428 6363943091 8325417290 0136503726 4357861146 5950013404 3088536429 5301770827 3894637360 4073219525 3363524736 8315637151 3409658626 2656344480 856171979[/code:1] Is there a formula of the exact x value now?[/code] 
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