A011974 : Is 2 the sum of 2 successive primes?

[zak seidov]

Thanks to all responding 
for idea about zeroth prime!

I've found TWO cases in OEIS, 
both give p(0)=1, p(1)=2, and if so 
the initial term of A011974'd be 3 not 2,
right?
zak  

Search: "zeroth prime" 
Displaying 1-2 of 2 results found. 
%N A059871 Number of solutions to the equation p_i =
(1+mod(i,2))*p_{i-1} +- p_{i-2} +- p_{i-3} 
               +- ... +- 2 +- 1, where p_i is the i-th
prime number (where p_1 = 
2, and the "zeroth prime" p_0 is here 1).

%C A062241 Here we are taking 1 to be the zeroth
prime.

--- zak seidov <zakseidov at yahoo.com> wrote:

> 
> A011974: Is 2 the sum of 2 successive primes? 
> zak
> 
> %I A011974
> %S A011974
> 2,5,8,12,18,24,30,36,42,52,60,68,78,84,90,100,112
> %N A011974 Numbers that are the sum of 2 successive
> primes.
> %D A011974 Archimedeans Problems Drive, Eureka, 26
> (1963), 12.
> %F A011974 Essentially same as A001043.
> 
> %I A001043 M3780 N0968
> %S A001043
> 5,8,12,18,24,30,36,42,52,60,68,78,84,90,100,112
> %N A001043 Numbers that are the sum of 2 successive
> primes.
> %D A001043 Archimedeans Problems Drive, Eureka, 26
> (1963), 12.
> 
> 
>      
>
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On Feb 13, 2008, Jonathan Post wrote:

> 2^0 + 3^1 + 5^2 + 7^3 + 11^4 = 15013 = Sum_{i=1..5) p(i)^(i-1), where
> p(i) denotes i-th prime number, is the largest known such prime (as of
> 13 Feb 2008)

It's always a little dangerous submitting a "largest" on the basis of  
a small list. This one is very easy to program and it takes only a few  

82630622945631213110177689366807452539856065087369361579159313857480591477585943781923604787789539375260939

...that is a probable prime: Sum(i=1..47) p(i)^(i-1)