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Search: id:A089577
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| A089577 |
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a(n) = r-th prime of the form (p-q)/(q-r) with r=prime(n+1), q=prime(n+2), and primes p>q. |
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| 13, 17, 47, 107, 179, 311, 401, 199, 761, 367, 1229, 1049, 1499, 701, 739, 1979, 1069, 2543, 2399, 1447, 3209, 1619, 4889, 4583, 4373, 4703, 4451, 5351, 5669, 6329, 2749, 6491, 5309, 7433, 4159, 3613, 8693, 4001, 3823, 9311, 7757, 10343, 11789, 10559, 5231, 6323, 14033, 12539, 14627, 6163, 13619, 12107, 7477, 7621, 7433, 15461, 7451, 17957, 16073, 14057, 18743, 20543, 19763, 22679, 20543, 10289
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Original definition: First occurrence of the first k-th occurrence where (p-q)/(q-k), p,q,k are primes.
From the original definition and program (which gives different results depending on the upper limit n) it is not obvious, but for any odd prime r, a solution will always be found for q=nextprime(r+1), by considering the r-th prime of the form p=q+(q-r)*prime(i), i=1,2,3... - M. F. Hasler, Apr 05 2009
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EXAMPLE
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For the first odd prime r=3 and its successor prime q=5, (p-q)/(q-r) is prime for p=11, 19, 31,... and the (r=3)-rd ratio is (31-5)/(5-3)=13=a(1). - M. F. Hasler, Apr 05 2009
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PROGRAM
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(PARI) diff2sqp2(n, r, d) = { forprime(q=d+1, n, c=0; forprime(p=q+1, n, y=(p-q)/(q-d); if(y==floor(y), if(isprime(y), c++; if(c==r, print1(y", "); break)) ) ) ) } diff2sqp3(n) = { forprime(k=3, n, diff2sqp2(n, k, k); print(); ) }
(PARI) A089577(n) = { local( q=prime(n+2), r=prime(n+1), d=q-r, y=1); until( isprime( q+d*y=nextprime(y+1)) & ! r--, ) ; y} \\ M. F. Hasler, Apr 05 2009
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CROSSREFS
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Sequence in context: A156553 A140533 A076789 this_sequence A060569 A108265 A069853
Adjacent sequences: A089574 A089575 A089576 this_sequence A089578 A089579 A089580
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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), Dec 29 2003
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EXTENSIONS
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Corrected, extended and edited by M. F. Hasler (MHasler(AT)univ-ag.fr), Apr 05 2009
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