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




Re: Sum of primes divided by the next prime
Posted:
Aug 2, 2006 4:58 PM


<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 >> > >



