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. :-)
<email@example.com> wrote in message news:firstname.lastname@example.org... > > 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 >