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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 4:58 PM
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<mensanator@aol.com> wrote in message
news:1154550547.288310.59840@i42g2000cwa.googlegroups.com...
>
> Tapio wrote:

>> 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.

>
> 2 is a solution to p_n but not a solution to p_(n+1).


:-)

Iff Sum (2+3+...+p_n)/(p_n) then also Sum (2+3+...+p_n)/(p_n+1) for p_n>2.

Tapio
>
>> :-)
>>
>> 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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