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Search: id:A136191
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| A136191 |
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Primes p such that 2p-3 and 2p+3 are both prime (A092110), with last decimal being 3. |
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5
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| 13, 43, 53, 113, 193, 223, 283, 563, 613, 643, 743, 773, 1033, 1193, 1453, 1483, 1543, 1583, 1663, 1733, 2143, 2393, 2503, 2843, 3163, 3413, 3433, 3793, 3823, 4133, 4463, 4483, 4523, 4603, 4673, 4813, 5443, 5743, 5953, 6073, 6133, 6163, 6553, 6733, 6863
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OFFSET
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1,1
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COMMENT
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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
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Cf. A092110, A136192.
Sequence in context: A118529 A129811 A108545 this_sequence A039318 A146507 A113831
Adjacent sequences: A136188 A136189 A136190 this_sequence A136192 A136193 A136194
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KEYWORD
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base,nonn
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AUTHOR
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Carlos Alves (cjsalves(AT)gmail.com), Dec 20 2007
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