

A114381


Sums of pth to the qth prime where p and q are consecutive primes.


5



8, 23, 41, 119, 109, 243, 187, 373, 689, 349, 991, 839, 551, 991, 1603, 1829, 841, 2155, 1717, 1079, 2689, 2081, 3113, 4359, 2641, 1667, 2867, 1779, 3037, 9905, 3627, 5293, 2357, 9125, 2599, 6265, 6593, 4889, 7081, 7327, 3219, 12253, 3487, 5933, 3631
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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



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 mth to the nth prime*/{ local(x); return(sum(x=m, n, prime(x))) }


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



