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Numbers k such that (10^k + 101)/3 is prime.
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%I #17 May 02 2024 22:54:29

%S 1,2,3,6,9,12,23,39,59,168,198,203,231,449,863,920,1064,1484,1674,

%T 2018,2943,3123,4073,4122,8360,11774,16031,26507,31146,33170,44952,

%U 62402,88020,89687

%N Numbers k such that (10^k + 101)/3 is prime.

%C For k > 1, numbers k such that k-2 occurrences of the digit 3 followed by the digits 67 is prime (see Example section).

%C a(35) > 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 3w67</a>.

%e 3 is in this sequence because (10^3+101)/3 = 367 is prime.

%e Initial terms and associated primes:

%e a(1) = 1, 37;

%e a(2) = 2, 67;

%e a(3) = 3, 367;

%e a(4) = 6, 333367;

%e a(5) = 9, 333333367, etc.

%t Select[Range[0, 100000], PrimeQ[(10^#+101)/3] &]

%o (PARI) lista(nn) = for(n=1, nn, if(ispseudoprime((10^n+101)/3), print1(n, ", "))); \\ _Altug Alkan_, Apr 16 2016

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

%K nonn,more

%O 1,2

%A _Robert Price_, Apr 16 2016