numerator of function is prime

[f.firoozbakht at sci.ui.ac.ir]

There is no known example where PrimeQ fails.
Mathematica says the numerator of f(n) for 

n=1,2,4,14,22,48,52,88,310,796  is prime.

Also if we define g(n)=sum(i=1,n,prime(i+1)/prime(i))
the numerator of g(n) for
n=1,2,3,4,7,32,53,55,94,183,189,404,480 
is prime(Mathematica says).

Firoozbakht 

Quoting Kennedy <kennedy at oldnews.org>:

> If pp(796) = f(796), where f is the sum you give,
> then Mathematica claims the numerator of pp(796)
> is prime. However, I seem to remember reading
> somewhere that Mathematica's PrimeQ predicate
> is not 100% reliable.
>  
> Kennedy
> 
> ----- Original Message ----- 
> From: Mohammed BOUAYOUN <mailto:Mohammed.BOUAYOUN at sanef.com>  
> To: seqfan at ext.jussieu.fr <mailto:seqfan at ext.jussieu.fr>  
> Sent: Monday, February 16, 2004 10:32 AM
> Subject: numerator of function is prime
> 
> I defined a new function f(n)=sum(i=1,n,prime(i)/prime(i+1))
> Example  
> f(1)=2/3
> f(2)=2/3 + 3/5
> f(3)=2/3+3/5+5/7
> f(4)=2/3+3/5+5/7+7/11
>  
> can anyone verify if numerator(pp(796)) is prime
>  
> thanks
> 
> 
> 


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