%I #10 Mar 20 2015 17:35:05
%S 1,1,1,1,1,3,3,1,3,2,3,4,1,2,2,3,2,1,2,3,4,3,2,1,3,2,4,5,2,7,6,7,11,6,
%T 6,3,7,7,8,11,10,3,4,6,10,4,3,4,3,3,4,6,4,4,4,4,3,6,4,3,6,6,5,7,5,11,
%U 5,7,8,4,4,7,7,7,10,3,6,10,2,1,6,4,6,3,4,3,1,5,4,4,5,6,3,6,1,4,3,4,6,3,5
%N Number of distinct primes embedded in prime p(n).
%C a(n) counts permuted subsequences of digits of p(n) which denote primes.
%C We put all the digits of prime(n) into a bag and ask how many distinct primes can be formed using some or all of these digits.
%e a(35)=6 because from the digits of p(35)=149, six numbers can be formed, 19, 41, 149, 419, 491 & 941, which are primes.
%t Needs["DiscreteMath`Combinatorica`"]; f[n_] := Length[ Union[ Select[ FromDigits /@ Flatten[ Permutations /@ Subsets[ IntegerDigits[ Prime[n]]], 1], PrimeQ]]]; Table[f[n], {n, 102}] (* _Ray Chandler_ and _Robert G. Wilson v_, Feb 25 2005 *)
%Y a(n) = A045719(n)+1 = A039993(p(n)) A101988 gives another version.
%K nonn,base
%O 1,6
%A _David W. Wilson_