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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.) 1
2, 3, 5, 7, 12, 21, 113, 115, 131, 151, 311, 511 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Comments from Adam M. Kalman (mocha(AT)clarityconnect.com), Nov 18 2004: (Start)

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).

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.

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:

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:

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.

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.

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)

LINKS

Table of n, a(n) for n=1..12.

Eric Weisstein's World of Mathematics, Symmetric Polynomial

EXAMPLE

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.

CROSSREFS

Cf. A046713, A086107.

Sequence in context: A318745 A062713 A334630 * A052430 A344454 A235153

Adjacent sequences:  A086105 A086106 A086107 * A086109 A086110 A086111

KEYWORD

full,fini,nonn,base

AUTHOR

Zak Seidov, Jul 10 2003

EXTENSIONS

Corrected by Adam M. Kalman (mocha(AT)clarityconnect.com), Nov 30 2004

STATUS

approved

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Last modified September 21 12:54 EDT 2021. Contains 347598 sequences. (Running on oeis4.)