%I #9 Jul 30 2019 17:16:50
%S 2,3,5,113,131,1511,111111113,1111111121,
%T 1111111111111111111111111111111121,
%U 111111111111111111111111111111111111131,111111111113111111111111111111111111111,111111131111111111111111111111111111111111111111111111111
%N Sophie Germain primes for which the product of the digits is also a Sophie Germain prime.
%C None of the numbers in the sequence can have digits 0,4,6,7,8 or 9. Either the digits are all 1's, or there is one digit 2,3 or 5 and all the others are 1's.
%C Comment from Hans Havermann, May 13 2006: If we express these numbers more compactly as (10^x-1)/9 + y*10^z, with y restricted to one of {1,2,4}, then the first 26 values (x < 2010) of {x,y,z} are: {1, 1, 0}, {1, 2, 0}, {1, 4, 0}, {3, 2, 0}, {3, 2, 1}, {4, 4, 2}, {9, 2, 0}, {10, 1, 1}, {34, 1, 1}, {39, 2, 1}, {39, 2, 27}, {57, 2, 49}, {82, 1, 39}, {114, 2, 84}, {129, 2, 69}, {142, 1, 132}, {148, 4, 119}, {148, 4, 132}, {160, 4, 53}, {160, 1, 105}, {244, 1, 16}, {280, 1, 210}, {976, 1, 285}, {1111, 1, 1000}, {1170, 2, 1094}, {1807, 1, 1308}.
%C The next term has 82 digits. - _Harvey P. Dale_, Jul 30 2019
%e 131 is in the sequence because (1) it is a Sophie Germain prime and (2) the product of its digits 1*3*1=3 is also a Sophie Germain prime.
%t Select[FromDigits/@(Flatten[Permutations/@Flatten[Table[PadRight[{n},k,1],{n,{1,2,3,5}},{k,60}],1],1]),AllTrue[ {#,2#+1,Times@@ IntegerDigits[ #],2Times@@ IntegerDigits[ #]+ 1},PrimeQ]&]//Sort (* _Harvey P. Dale_, Jul 30 2019 *)
%Y Cf. A005384.
%K base,nonn
%O 1,1
%A Luc Stevens (lms022(AT)yahoo.com), May 06 2006
%E More terms from _Hans Havermann_, May 07 2006