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%I #14 Nov 23 2019 04:12:45
%S 0,1,3,6,10,12,16,19,21,22,27,33,34,36,40,45,48,51,54,58,61,70,72,85,
%T 87,90,94,96,103,105,106,111,112,118,121,124,126,127,133,135,136,139,
%U 147,148,150,153,154,159,177,180,183,184,187,189,190,192,198,199,201,210,213,216
%N Numbers n such that 100n + 13 is prime.
%C The sequence is infinite, because by Dirichlet's theorem there are infinitely many primes in the arithmetic sequence A*n+B (n=1,2,...) if A an B are relatively prime.
%C The sequence also has an infinite set of pairs a(k+1)=a(k)+1 (two consecutive naturals), but no set of three consecutive naturals (each third natural is divisible by 3)
%C No term of the sequence is of form 3k+2, because the sum of digits of 100*(3k+2)+13 is divisible by 3, violating the requirement of the definition.
%C Indices (as k-th prime) of the first members are 6, 30, 65, 112, 170, 198, 255, 293, 319, 330, 396, 466, 480, 505, 554, 612, 648, 684, 714, 763, 797, 902, 922, 1061, 1086, 1121, 1164, 1186, 1265, 1286, 1295, ...
%F {a(n): 100*a(n)+13 in A000040}.
%e a(1)=0: 100*0+13=13 smallest prime which ends in 13, see A000040(6).
%e a(2)=1: 100*1+13=113 second prime which ends in 13, see A000040(30).
%o (PARI) isok(n) = isprime(100*n + 13) \\ _Michel Marcus_, Jul 22 2013
%Y Cf. A088262 (6th row). - _R. J. Mathar_, Apr 18 2009
%K nonn,easy
%O 1,3
%A Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 06 2009
%E Edited, 27 inserted by _R. J. Mathar_, Apr 18 2009
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