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Topic: Sum of primes divided by the next prime
Replies: 6   Last Post: Aug 2, 2006 4:58 PM

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Tapio Hurme

Posts: 853
Registered: 12/8/04
Re: Sum of primes divided by the next prime
Posted: Aug 2, 2006 3:29 PM
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Thanks for the correction! My typo. :-(
So, the problem is the same as: Sum (2+3+...+p_n)/(p_n).

Hmm...What we can learn about the next prime p_(n+1) versus p_n.
:-)

Tapio


<jankrihau@hotmail.com> wrote in message
news:1154546221.477259.213700@s13g2000cwa.googlegroups.com...
>
> Tapio wrote:

>> Sum the primes up to p_n. Divide the sum by the next prime p_(n+1):
>>
>> Sum (2+3+...+p_n)/(p_(n+1))
>>
>> I found with the aid of excel two examples:
>>
>> (2+3) = 0 mod 5, i.e. the next prime 5 divides the sum of all the
>> preceding
>> primes.
>>
>> (2+3+....+61) = 0 mod 67

>
> This is in error; 2+3+...+67 is divisible by 71.
>

>> Below the prime 10009 there is no other cases, except those mentioned
>> above.
>> Are there any other p_n > 10009 ?

>
> Check this link:
>
> http://www.research.att.com/~njas/sequences/A007506
>
> ---
> J K Haugland
> http://home.no.net/zamunda
>






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