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A114381
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Sums of pth to the qth prime where p and q are consecutive primes.
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4
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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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. [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jul 01 2009]
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LINKS
| Klaus Brockhaus, Table of n, a(n) for n=1..3245. [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jul 01 2009]
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FORMULA
| prime(n) is the n-th prime number.
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EXAMPLE
| 7 and 11 are consecutive primes. prime(7)+prime(8)+prime(9)+prime(10)+prime(11)=
119, the 4th entry in the table.
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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))) }
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CROSSREFS
| A161926 (numbers occurring at least twice), A161927 (index of second occurrence). [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jul 01 2009]
Sequence in context: A164284 A047719 A164131 * A139433 A178072 A185257
Adjacent sequences: A114378 A114379 A114380 * A114382 A114383 A114384
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KEYWORD
| easy,nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)gmail.com), Feb 10 2006
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