%I #7 Mar 31 2012 14:40:08
%S 85,145,304,915
%N Numbers n such that 2111131111...p(n-1)1111p(n) is prime.
%C Let fp(n,k) be the decimal concatenation of prime(n), k, and fp(n-1,k) for n > 1, and fp(1,k) = 2. Then a(n) = fp(n, 1111).
%C No more terms up to 6300.
%H Farideh Firoozbakht, <a href="http://www.primepuzzles.net/puzzles/puzz_208.htm">On Solution of puzzle 208 </a>.
%e a(1)=85 because fp(85,1111)= 211113111151111...4331111439 is prime and fp(k,1111) is composite for k< 85 (prime(85)=439).
%e a(3)=304 because fp(304,1111)=211113111151111...199911112003 is a prime related to prime year 2003; this prime number has 2231 digits. fp(915,1111)=211113111151111...712911117151 is a prime with 7119 digits (prime(915)=7151).
%Y Cf. A083677, A082549.
%K nonn,base
%O 1,1
%A _Farideh Firoozbakht_, Jun 17 2003
%E Comment from _Charles R Greathouse IV_, Oct 12 2009
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