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A112130
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Numbers k such that (3^j)*k + 1 are primes for j=0 to 7.
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2
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25451790, 29445850, 76355370, 218715490, 236862990, 380011170, 514144750, 628241740, 777146230, 882792120, 930646080, 944173860, 1105472340, 1349221230, 1542434250, 1564227910, 1832212270, 1898927100, 1994085030
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OFFSET
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1,1
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COMMENTS
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Each term is a multiple of 70. The proof is straightforward. Example step showing k <> 3 (mod 7): If k==3 (mod 7), then (3^2)*k+1 == 9*3+1 == 0 (mod 7); i.e., for j=2 (3^j)*k+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)*k+1 is composite (a multiple of that modulus) so that only k == 0 (mod 2), k == 0 (mod 5) and k == 0 (mod 7) remain, hence k == 0 (mod 70). - Rick L. Shepherd, Sep 03 2005
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LINKS
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MATHEMATICA
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With[{c=3^Range[0, 7]}, Select[70*Range[3*10^7], AllTrue[1+c #, PrimeQ]&]] (* Harvey P. Dale, Sep 06 2023 *)
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PROG
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(PARI) forstep(k=70, 3*10^9, 70, j=0; while(isprime((3^j)*k+1), j++); if(j>=8, print1(k, ", "))); \\ Rick L. Shepherd, Sep 03 2005
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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