%I #14 Jun 26 2018 04:56:03
%S 1,1,4,7,7,170716,170926,26373004,247201983,10562770680,118345066231,
%T 54717848613610
%N a(n) is the smallest k such that 10^m*k + 1 is prime for all m in 1..n.
%C a(12) > 3*10^11.
%e 10^1*1 + 1 = 11 (prime), so a(1) = 1.
%e 10^2*1 + 1 = 101 (also prime), so a(2) = 1 as well.
%e 10^3*1 + 1 = 1001 = 7*143, so a(3) > 1;
%e 10^1*2 + 1 = 21 = 3*7, so a(3) > 2;
%e 10^2*3 + 1 = 301 = 7*43, so a(3) > 3;
%e however, for m = 1..3, 10^m*4 + 1 yields 41, 401, and 4001, each of which is prime, so a(3) = 4.
%Y Cf. A000040 (primes), A124417, A124516.
%K nonn,hard,more
%O 1,3
%A _Jon E. Schoenfield_, Jun 23 2018
%E a(12) from _Giovanni Resta_, Jun 25 2018