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Search: id:A114381
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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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+0
1
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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; listen)
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OFFSET
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1,1
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COMMENT
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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.
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FORMULA
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prime(n) is the n-th prime number.
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EXAMPLE
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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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PROGRAM
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(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
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Sequence in context: A034811 A003342 A047719 this_sequence A139433 A033951 A027054
Adjacent sequences: A114378 A114379 A114380 this_sequence A114382 A114383 A114384
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KEYWORD
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easy,nonn
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AUTHOR
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Cino Hilliard (hillcino368(AT)gmail.com), Feb 10 2006
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