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Search: id:A133313
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| A133313 |
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Primes p such that 3p-2 and 3p+2 are primes (see A125272) and its decimal representation finishes with 3. |
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1
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| 3, 13, 23, 43, 103, 163, 293, 313, 433, 523, 953, 1013, 1063, 1153, 1283, 1303, 1483, 1693, 1723, 1783, 1913, 2003, 2333, 3533, 3823, 3943, 4003, 4013, 4093, 4943, 5483, 6043, 6133, 6173, 6473, 6803, 7523, 7573, 7603, 7673, 7853, 7993, 8513, 9283, 9343
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OFFSET
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1,1
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COMMENT
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Theorem: If in the triple (3n-2,n,3n+2) all numbers are primes, then n=5 or the decimal representation of n finishes with 3 or 7. Proof: Similar to A136191. Alternative Mathematica proof: Table[nn = 10k + r; Intersection (AT)(AT) (Divisors[CoefficientList[(3nn - 2) nn(3nn + 2), k]]), {r, 1, 9, 2}]; This gives {{1, 5}, {1}, {1, 5}, {1}, {1, 5}}. Therefore only r=3 and r=7 allow non trivial divisors (excluding nn=5 itself).
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MATHEMATICA
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TPrimeQ = (PrimeQ[ # - 2] && PrimeQ[ #/3] && PrimeQ[ # + 2]) &; Select[Select[Range[100000], TPrimeQ]/3, Mod[ #, 10] == 3 &]
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CROSSREFS
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Cf. A136204 (finishing with 7), A136191, A136192, A125272.
Adjacent sequences: A133310 A133311 A133312 this_sequence A133314 A133315 A133316
Sequence in context: A121756 A030431 A090146 this_sequence A102010 A121718 A030552
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KEYWORD
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nonn,base
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AUTHOR
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Carlos Alves (cjsalves(AT)gmail.com), Dec 21 2007
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