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A136192
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Primes p such that 2p-3 and 2p+3 are both prime (A092110), with last decimal being 7.
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4
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7, 17, 67, 97, 127, 137, 157, 167, 487, 547, 617, 647, 937, 1187, 1277, 1427, 1627, 1847, 2027, 2297, 2437, 2467, 2477, 2617, 2857, 2927, 3137, 3457, 3727, 4007, 4057, 4157, 5167, 5417, 5657, 6247, 6257, 7027, 7477, 7867, 8467, 8737, 8747, 9127, 9227
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Except for p=5, the decimals in A092110 end in 3 or 7.
Theorem: If in the triple (2n-3,n,2n+3) all numbers are primes then n=5 or the decimal representation of n ends in 3 or 7. Proof: Consider Q=(2n-3)n(2n+3), by hypothesis factorized into primes. If n is prime, n=10k+r with r=1,3,7 or 9. We want to exclude r=1 and r=9. Case n=10k+1. Then Q=5(-1+6k+240k^2+800k^3) and 5 is a factor; thus 2n-3=5 or n=5 or 2n+1=5 : this means n=4 (not prime); or n=5 (included); or n=2 (impossible, because 2n-3=1). Case n=10k+9. Then Q=5(567+1926k+2160k^2+800k^3) and 5 is a factor; the arguments, for the previous case, also hold.
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CROSSREFS
| Cf. A092110, A136191.
Sequence in context: A094534 A081632 A106010 * A118431 A051809 A034054
Adjacent sequences: A136189 A136190 A136191 * A136193 A136194 A136195
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KEYWORD
| nonn,base
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AUTHOR
| Carlos Alves (cjsalves(AT)gmail.com), Dec 20 2007
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