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Numbers k such that (4*10^k + 143) / 3 is prime.
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%I #17 Jun 02 2024 09:28:43

%S 1,2,3,4,7,9,15,21,22,44,49,53,63,127,145,393,856,1006,1883,2263,5684,

%T 13324,14291,27435,38897,114076

%N Numbers k such that (4*10^k + 143) / 3 is prime.

%C For k > 1, numbers k such that the digit 1 followed by k-2 occurrences of the digit 3 followed by the digits 81 is prime (see Example section).

%C a(27) > 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 13w81</a>.

%e 2 is in this sequence because (4*10^2 + 143) / 3 = 1381 is prime.

%e Initial terms and associated primes:

%e a(1) = 1, 61;

%e a(2) = 2, 181;

%e a(3) = 3, 1381;

%e a(4) = 4, 13381;

%e a(5) = 7, 13333381, etc.

%t Select[Range[0, 100000], PrimeQ[(4*10^# + 143) / 3] &]

%o (PARI) is(n) = ispseudoprime((4*10^n + 143) / 3); \\ _Altug Alkan_, Sep 20 2016

%o (Magma) [n: n in [0..500] | IsPrime((4*10^n+143) div 3)]; // _Vincenzo Librandi_, Sep 22 2016

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

%K nonn,more

%O 1,2

%A _Robert Price_, Sep 20 2016

%E a(26) from _Robert Price_, Mar 05 2018