International Mathematical Forum, 5, 2010, no. 27, 1303 - 1322 On Symmetric Polynomials and Number Theory Rafael Jakimczuk Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina [email protected] Abstract It is well known that the symmetric function xk1xk2 ...xkh 1 2 h in n indeterminates x1,x2,...,xn may be expressed as a polynomial in terms of the more simple symmetric functions xk xk ... xk 1 + 2 + + n In this article we obtain some results on these polynomials. Finally, we apply these results to some problems in number theory on sums of products of positive integers. For example, let A be the set of the first n primes, that is A = {p1,p2,...,pn}. Let An be the sum of all possible products of k different primes in A. We prove k k k 2k k (p1 + p2 + ...+ pn) n pn n log n An ∼ ∼ ∼ k! k!2k k!2k Mathematics Subject Classification: 11B99, 11N45 Keywords: Symmetric polynomials, sums of products of positive integers, formulas, asymptotic formulas 1 Symmetric Polynomials A polynomial in the indeterminates x1,x2,...,xn which is unchanged by any permutation of the indeterminates is called a symmetric polynomial. The following symmetric polynomials are very important in the theory of these polynomials. 1304 R. Jakimczuk Definition 1.1 Given the indeterminates x1,x2,...,xn, let us consider the set of all possible monomials such that they have L (L ≥ 1) different indetermi- nates, t (t ≥ 1) indeterminates, and furthermore L1 different indeterminates have exponent K1, L2 different indeterminates have exponent K2 , ..., Lm different indeterminates have exponent Km (K1 >K2 >...> Km). The sum of all these monomials with coefficient 1 is a polynomial in the indeterminates x1,x2,...,xn which we shall denote (L1,K1 : L2,K2 : ...: Lm,Km) (1) Note that L1 + L2 + ...+ Lm = L (2) L1K1 + L2K2 + ...+ LmKm = t (3) Furthermore, clearly the number of terms (or monomials) in this polynomial is n(n − 1) ...(n − (L − 1)) (n ≥ L) (4) L1!L2! ...Lm! Example 1.2 If the indeterminates are x1,x2,x3,x4 we have , x2 x2 x2 x2 (1 2) = 1 + 2 + 3 + 4 (2, 1) = x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4 (4, 1) = x1x2x3x4 , , , x3x3x2x x3x3x2x x3x3x2x x3x3x2x x3x3x2x (2 3:1 2:1 1) = 1 2 3 4 + 1 2 4 3 + 1 3 2 4 + 1 3 4 2 + 1 4 2 3 x3x3x2x x3x3x2x x3x3x2x x3x3x2x x3x3x2x + 1 4 3 2 + 2 3 1 4 + 2 3 4 1 + 2 4 1 3 + 2 4 3 1 x3x3x2x x3x3x2x + 3 4 1 2 + 3 4 2 1 Consider the polynomials of the form, H ,r xr xr ... xr r =(1 )= 1 + 2 + + n It is well known that the polynomial (1) may be expressed as a polynomial of rational coefficients in terms of the Hr where r takes certain values. That is P (Hi ,Hi ,...,Hi ) L ,K L ,K ... L ,K 1 2 k ( 1 1 : 2 2 : : m m)= A (5) P H ,H ,...,H H ,H ,...,H Where ( i1 i2 ik ) is a polynomial of integer coefficients in i1 i2 ik and A is a positive integer. The following combinatorial general formulas (L ≥ 2) will be our key lemma. Symmetric polynomials and number theory 1305 Lemma 1.3 If L1 ≥ 2, L1(L1,K1 : L2,K2 : ...: Lm,Km) H L − ,K L ,K ... L ,K = K1 ( 1 1 1 : 2 2 : : m m) − (1,K1 + K1 : L1 − 2,K1 : L2,K2 : ...: Lm,Km) − (1,K1 + K2 : L1 − 1,K1 : L2 − 1,K2 : ...: Lm,Km) − ... − (1,K1 + Km : L1 − 1,K1 : L2,K2 : ...: Lm − 1,Km) (6) If L1 =1, (1,K1 : L2,K2 : ...: Lm,Km) H L ,K ... L ,K = K1 ( 2 2 : : m m) − (1,K1 + K2 : L2 − 1,K2 : ...: Lm,Km) − ... − (1,K1 + Km : L2,K2 : ...: Lm − 1,Km) (7) Proof. These formulas can be proved without difficulty from definition 1.1. See for example [6] (page 82, exercise 5), where a similar formula more complicated is established. Our formulas are more simple since we assume K1 >K2 >...> Km (see definition 1.1). Remark 1. Note that the polynomial in the left side of (6) or (7) has L different indeterminates in its monomials (see definition 1.1). On the other hand, the polynomials in the right side of (6) or (7) have L − 1 different indeterminates in their monomials. This fact is important in the proof by mathematical induction on L of a set of properties of these polynomials (see below). Remark 2. Note that the polynomial in the left side of (6) or (7) has t indeterminates in its monomials (see definition 1.1). Note that (see (6)) the polynomial (L1 −1,K1 : L2,K2 : ...: Lm,Km) has t−K1 indeterminates in its monomials and the m polynomials with sign minus have t indeterminates in their monomials. Analogously (see (7)) the polynomial (L2,K2 : ...: Lm,Km) has t − K1 indeterminates in its monomials and the m − 1 polynomials with sign minus have t indeterminates in their monomials. If we apply (6) or (7) repeatedly, clearly we obtain the polynomial (5). Since in each application L decreases in one and if L = 2 formulas (7) and (6) give ,K ,K H ,K − ,K K H H − H (1 1 :1 2)= K1 (1 2) (1 1 + 2)= K1 K2 K1+K2 (8) ,K H ,K − ,K K H2 − H 2(2 1)= K1 (1 1) (1 1 + 1)= K1 K1+K1 (9) 1306 R. Jakimczuk Example 1.4 a) Given the indeterminates x1,x2,...,xn consider the poly- nomial (2, 2:1, 1) where L =3. Formula (6) gives 2(2, 2:1, 1) = H2(1, 2:1, 1) − (1, 4:1, 1) − (1, 3:1, 2:0, 1) = H2(1, 2:1, 1) − (1, 4:1, 1) − (1, 3:1, 2) (10) Formula (8) gives (1, 2:1, 1) = H2H1 − H3 (1, 4:1, 1) = H4H1 − H5 (1, 3:1, 2) = H3H2 − H5 Substituting these formulas into (10) we obtain the polynomial (5) that corre- spond to (2, 2:1, 1) P (H ,H ,H ,H ,H ) H2H − 2H H − H H +2H (2, 2:1, 1) = 1 2 3 4 5 = 2 1 3 2 4 1 5 (11) 2 2 b) Given the indeterminates x1,x2,...,xn consider the polynomial (2, 3:1, 2:1, 1) where L =4. Formula (6) gives 2(2, 3:1, 2:1, 1) = H3(1, 3:1, 2:1, 1) − (1, 6:1, 2:1, 1) − (1, 5:1, 3:1, 1) − (1, 4:1, 3:1, 2) (12) Formula (7) gives (1, 3:1, 2:1, 1) = H3(1, 2:1, 1) − (1, 5:1, 1) − (1, 4:1, 2) (1, 6:1, 2:1, 1) = H6(1, 2:1, 1) − (1, 8:1, 1) − (1, 7:1, 2) (1, 5:1, 3:1, 1) = H5(1, 3:1, 1) − (1, 8:1, 1) − (1, 6:1, 3) (1, 4:1, 3:1, 2) = H4(1, 3:1, 2) − (1, 7:1, 2) − (1, 6:1, 3) (13) Formula (8) gives (1, 2:1, 1) = H2H1 − H3 (1, 5:1, 1) = H5H1 − H6 (1, 4:1, 2) = H4H2 − H6 (1, 8:1, 1) = H8H1 − H9 (1, 7:1, 2) = H7H2 − H9 Symmetric polynomials and number theory 1307 (1, 3:1, 1) = H3H1 − H4 (1, 6:1, 3) = H6H3 − H9 (1, 3:1, 2) = H3H2 − H5 (14) Substituting (14) into (13) we obtain , , , H H H − H2 − H H − H H H (1 3:1 2:1 1) = 3 2 1 3 5 1 4 2 +2 6 (1, 6:1, 2:1, 1) = H6H2H1 − H6H3 − H8H1 − H7H2 +2H9 (1, 5:1, 3:1, 1) = H5H3H1 − H5H4 − H8H1 − H6H3 +2H9 (1, 4:1, 3:1, 2) = H4H3H2 − H5H4 − H7H2 − H6H3 +2H9 (15) Finally, substituting (15) into (12) we obtain the polynomial (5) that corre- spond to (2, 3:1, 2:1, 1) , , , H2H H − H3 − H H H − H H H − H H H (2 3:1 2:1 1) = 3 2 1 3 2 5 3 1 2 4 3 2 6 2 1 +5H6H3 +2H8H1 +2H7H2 +2H5H4 − 6H9) /2 (16) c) We have (formula (6)) 2(2, 1) = H1(1, 1) − (1, 2) = H1H1 − H2 (17) We have (formulas (6) and (7)) 3(3, 1) = H1(2, 1) − (1, 2:1, 1) (1, 2:1, 1) = H2H1 − H3 Therefore 3(3, 1) = H1(2, 1) − H2H1 + H3 (18) We have (formulas (6) and (7)) 4(4, 1) = H1(3, 1) − (1, 2:2, 1) (1, 2:2, 1) = H2(2, 1) − (1, 3:1, 1) (1, 3:1, 1) = H3H1 − H4 Therefore 4(4, 1) = H1(3, 1) − H2(2, 1) + H3H1 − H4 (19) In this way we obtain the general recursive formula k k+1 k(k, 1) = H1(k − 1, 1) − H2(k − 2, 1) + ...+(−1) Hk−1H1 +(−1) Hk (20) 1308 R. Jakimczuk where k ≥ 2. (17) gives H2 − H (2, 1) = 1 2 (21) 2! (18) and (21) give H3 − 3H H +2H (3, 1) = 1 2 1 3 (22) 3! (19), (21) and (22) give H4 − 6H H2 +8H H +3H2 − 6H (4, 1) = 1 2 1 3 1 2 4 (23) 4! (20) with k =5, (23), (22) and (21) give H5 − 10H H3 +20H H2 +15H 2H − 30H H − 20H H +24H (5, 1) = 1 2 1 3 1 2 1 4 1 3 2 5 5! (24) . From (6), (7), (8) and (9) can be proved by mathematical induction on L the following properties. These properties are our main results in this section.