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Search: id:A112130
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| A112130 |
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Numbers n such that (3^j)*n + 1 are primes for j=0 to 7. |
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+0
2
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| 25451790, 29445850, 76355370, 218715490, 236862990, 380011170, 514144750, 628241740, 777146230, 882792120, 930646080, 944173860, 1105472340, 1349221230, 1542434250, 1564227910, 1832212270, 1898927100, 1994085030
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Part of sequence A112129
Each term is a multiple of 70. The proof is straightforward. Example step showing n <> 3 (mod 7): If n==3 (mod 7), then (3^2)*n+1 == 9*3+1 == 0 (mod 7); i.e., for j=2 (3^j)*n+1 is never prime in this case. A corresponding j value with 0<=j<=7 can be found for each modulus (2,5,7) and nonzero residue such that (3^j)*n+1 is composite (a multiple of that modulus) so that only n==0 (mod 2), n==0 (mod 5) and n==0 (mod 7) remain, hence n==0 (mod 70). - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Sep 03 2005
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PROGRAM
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(PARI) forstep(n=70, 3*10^9, 70, j=0; while(isprime((3^j)*n+1), j++); if(j>=8, print1(n, ", "))) (Shepherd)
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CROSSREFS
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Cf. A112129.
Sequence in context: A083636 A124069 A056915 this_sequence A120702 A113026 A120410
Adjacent sequences: A112127 A112128 A112129 this_sequence A112131 A112132 A112133
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KEYWORD
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nonn
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AUTHOR
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Pierre CAMI (pierrecami(AT)tele2.fr), Aug 27 2005
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EXTENSIONS
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More terms from Rick L. Shepherd (rshepherd2(AT)hotmail.com), Sep 03 2005
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