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%I #13 Sep 06 2023 16:07:52
%S 25451790,29445850,76355370,218715490,236862990,380011170,514144750,
%T 628241740,777146230,882792120,930646080,944173860,1105472340,
%U 1349221230,1542434250,1564227910,1832212270,1898927100,1994085030
%N Numbers k such that (3^j)*k + 1 are primes for j=0 to 7.
%C 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
%H Jinyuan Wang, <a href="/A112130/b112130.txt">Table of n, a(n) for n = 1..500</a>
%t With[{c=3^Range[0,7]},Select[70*Range[3*10^7],AllTrue[1+c #,PrimeQ]&]] (* _Harvey P. Dale_, Sep 06 2023 *)
%o (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
%Y Subsequence of A112129.
%K nonn
%O 1,1
%A _Pierre CAMI_, Aug 27 2005
%E More terms from _Rick L. Shepherd_, Sep 03 2005
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