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Numbers k such that (5*10^k + 79)/3 is prime.
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%I #18 May 11 2024 19:24:32

%S 1,2,3,4,5,8,9,20,291,417,712,749,1906,2086,2746,3896,4927,10058,

%T 18369,34071,36569,44749,89510,139457

%N Numbers k such that (5*10^k + 79)/3 is prime.

%C For k > 1, numbers k such that the digit 1 followed by k-2 occurrences of the digit 6 followed by the digits 93 is prime (see Example section).

%C a(25) > 2*10^5.

%H Makoto Kamada, <a href="https://stdkmd.net/nrr">Factorization of near-repdigit-related numbers</a>.

%H Makoto Kamada, <a href="https://stdkmd.net/nrr/prime/prime_difficulty.txt">Search for 16w93</a>.

%e 2 is in this sequence because (5*10^2 + 79)/3 = 193 is prime.

%e Initial terms and associated primes:

%e a(1) = 1, 43;

%e a(2) = 2, 193;

%e a(3) = 3, 1693;

%e a(4) = 4, 16693;

%e a(5) = 5, 166693; etc.

%t Select[Range[0, 100000], PrimeQ[(5*10^# + 79)/3] &]

%o (PARI) is(k) = ispseudoprime((5*10^k + 79)/3) \\ _Iain Fox_, Nov 12 2017

%Y Cf. A056654, A268448, A269303, A270339, A270613, A270831, A270890, A270929, A271269.

%K nonn,more,hard

%O 1,2

%A _Robert Price_, Nov 12 2017

%E a(24) from _Robert Price_, Mar 09 2018