%I A086108
%S A086108 2,3,5,7,12,21,113,115,131,151,311,511
%N A086108 Numbers n with the following property: Every symmetric polynomial of 
               the digits of n is prime. (A symmetric polynomial is unchanged by 
               any permutation of its variables, so the symmetric polynomials of 
               {a,b,c} would be a+b+c,ab+bc+ac and abc.).
%C A086108 Comments from Adam M. Kalman (mocha(AT)clarityconnect.com), Nov 18 2004: 
               "The k-th symmetric polynomial of {a1,a2,a3,...an} can also be thought 
               of as the coefficient of x^(n-k) in the binomial expansion of (x-a1)(x-a2)...(x-an).
%C A086108 "Derivation of full sequence: First, no member of this sequence can contain 
               a composite digit, or else the last symmetric polynomial (the product 
               of its digits) will be composite. Second, no member may contain more 
               than one prime digit, for the same reason. Third, any member will 
               have all permutations of its digits also in the sequence.
%C A086108 "Therefore in order to find members of this sequence, we need only examine 
               the sets of digits {1,1,1,...,1,p}, where p is either 2,3,5, or 7 
               (the prime digits) and there are n ones in the set. In the cases 
               of these sets, it is easy to see what the symmetric polynomials are:
%C A086108 "The number of times p appears in the k-th symmetric polynomial is binomial[n,
               k-1] and the number of times p doesn't appear is binomial[n,k]. Therefore 
               the k-th symmetric polynomial of this set is p*binomial[n,k-1]+binomial[n,
               k]. But now consider the sets with n>2. Observe that the second symmetric 
               polynomial is given by substituting k=2 into the above formula:
%C A086108 "q = p*binomial[n,1]+binomial[n,2] = pn+(n)(n-1)/2. If n is even, then 
               q can be factored nontrivially into integers: (n/2)(2p+n-1). If n 
               is odd, then q can be factored nontrivially as well: (n)(p+(n-1)/
               2). Therefore in these cases q (the second symmetric polynomial) 
               is always composite and so no set with n>2 (i.e. containing more 
               than 2 ones) can have the desired property.
%C A086108 "This means that we only have to examine sets with 0, 1 and 2 ones and 
               so we immediately see that the sequence is finite and short. Furthermore, 
               examining these 12 sets ({2},{3},{5},{7},{1,2},{1,3},{1,5},{1,7},
               {1,1,2},{1,1,3},{1,1,5},{1,1,7}), we immediately eliminate sets whose 
               sums (the first symmetric polynomial) are composite, leaving only 
               ({2},{3},{5},{7},{1,2},{1,1,3},{1,1,5}) for consideration.
%C A086108 "A minute of calculation shows that these seven sets all have the desired 
               property and so the full sequence consists of all integers whose 
               digits are permutations of those seven sets: 2,3,5,7,12,21,113,115,
               131,151,311,511."
%H A086108 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               SymmetricPolynomial.html">SymmetricPolynomial</a>
%e A086108 The number 131 is in the sequence because every symmetric polynomial 
               of {1,3,1} is prime: 1+3+1=5, 1*3+3*1+1*1=7 and 1*3*1=3 are all prime.
%Y A086108 Cf. A046713, A086107.
%Y A086108 Sequence in context: A018065 A048818 A062713 this_sequence A052430 A024784 
               A060528
%Y A086108 Adjacent sequences: A086105 A086106 A086107 this_sequence A086109 A086110 
               A086111
%K A086108 full,fini,nonn,base
%O A086108 1,1
%A A086108 Zak Seidov (zakseidov(AT)yahoo.com), Jul 10 2003
%E A086108 Corrected by Adam M. Kalman (mocha(AT)clarityconnect.com), Nov 30 2004