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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.)

%I A086108 #18 Feb 16 2025 08:32:50
%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: (Start)
%C A086108 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. (End)
%H A086108 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SymmetricPolynomial.html">Symmetric Polynomial</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, A052430.
%K A086108 full,fini,nonn,base
%O A086108 1,1
%A A086108 _Zak Seidov_, Jul 10 2003
%E A086108 Corrected by Adam M. Kalman (mocha(AT)clarityconnect.com), Nov 30 2004