OFFSET
1,1
COMMENTS
The number of terms in this sequence is infinite since there is no largest prime number. Conjecture: There will always be an n and i such that a(n) >= a(n+i) or the sequence will alternate forever. Equality does take place in the small sample shown with the entry 991. Certainly the proof of an infinity many twin primes would be a strong probable proof of this assertion. My guess is the alternation would always occur when a twin prime is encountered and often for other consecutive primes such as those differing by 4.
Some numbers occur (at least) twice: 991 at positions 11 and 14, 104435 at positions 193 and 348, 712363 at positions 654 and 2364. - Klaus Brockhaus, Jul 01 2009
LINKS
Klaus Brockhaus, Table of n, a(n) for n=1..3245.
FORMULA
a(n) = Sum_{k=prime(n)..prime(n+1)} prime(k). - Danny Rorabaugh, Apr 01 2015
EXAMPLE
7 and 11 are consecutive primes. prime(7)+prime(8)+prime(9)+prime(10)+prime(11)= 119, the 4th entry in the table.
PROG
(PARI) g2(n)=for(x=1, n, print1(sumprimes(prime(x), prime(x+1))", "))
sumprimes(m, n) = /* Return the sum of the m-th to the n-th prime*/{ local(x); return(sum(x=m, n, prime(x))) }
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, Feb 10 2006
STATUS
approved