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Numbers k such that (299*10^k - 17)/3 is prime.
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%I #21 May 26 2024 15:24:07

%S 1,3,6,12,16,18,27,30,59,60,61,118,198,208,826,1696,1813,4505,7111,

%T 9715,11572,15439,17406,55998,89836,158544,199801,201547

%N Numbers k such that (299*10^k - 17)/3 is prime.

%C For k > 1, numbers k such that the digits 99 followed by k-1 occurrences of the digit 6 followed by the digit 1 is prime (see Example section).

%C a(29) > 3*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 996w1</a>.

%e 3 is in this sequence because (299*10^3 - 17) / 3 = 99661 is prime.

%e Initial terms and associated primes:

%e a(1) = 1, 991;

%e a(2) = 3, 99661;

%e a(3) = 6, 99666661;

%e a(4) = 12, 99666666666661;

%e a(5) = 16, 996666666666666661; etc.

%t Select[Range[0, 100000], PrimeQ[(299*10^# - 17) / 3] &]

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

%K nonn,more,hard

%O 1,2

%A _Robert Price_, Jan 13 2017

%E a(26)-a(27) from _Robert Price_, Apr 15 2020

%E Constant 229 corrected to 299 by _Georg Fischer_, Jun 26 2020

%E a(28) from _Robert Price_, Jun 21 2023