On Symmetric Polynomials and Number Theory 1 Symmetric

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we apply these results to some problems in number theory on sums of products of ..... Note that if the number of indeterminates xi is L−1 then the right side in (6).
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 

k2 kh xk1 1 x2 . . . xh

in n indeterminates x1 , x2 , . . . , xn may be expressed as a polynomial in terms of the more simple symmetric functions xk1 + xk2 + . . . + xkn 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 An ∼

nk pkn (p1 + p2 + . . . + pn )k n2k logk n ∼ ∼ 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.

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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 indeterminates, 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)

L1 + L2 + . . . + Lm = L

(2)

L1 K1 + L2 K2 + . . . + Lm Km = t

(3)

Note that

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 (1, 2) = x21 + x22 + x23 + x24 (2, 1) = x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4 (4, 1) = x1 x2 x3 x4 (2, 3 : 1, 2 : 1, 1) = x31 x32 x23 x4 + x31 x32 x24 x3 + x31 x33 x22 x4 + x31 x33 x24 x2 + x31 x34 x22 x3 + x31 x34 x23 x2 + x32 x33 x21 x4 + x32 x33 x24 x1 + x32 x34 x21 x3 + x32 x34 x23 x1 + x33 x34 x21 x2 + x33 x34 x22 x1 Consider the polynomials of the form, Hr = (1, r) = xr1 + xr2 + . . . + xrn 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 (L1 , K1 : L2 , K2 : . . . : Lm , Km ) =

P (Hi1 , Hi2 , . . . , Hik ) A

(5)

Where P (Hi1 , Hi2 , . . . , Hik ) is a polynomial of integer coefficients in Hi1 , Hi2 , . . . , Hik and A is a positive integer. The following combinatorial general formulas (L ≥ 2) will be our key lemma.

Symmetric polynomials and number theory

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Lemma 1.3 If L1 ≥ 2,

= − − −

L1 (L1 , K1 : L2 , K2 : . . . : Lm , Km ) HK1 (L1 − 1, K1 : L2 , K2 : . . . : Lm , Km ) (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 ) = HK1 (L2 , K2 : . . . : Lm , Km ) − (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 (1, K1 : 1, K2 ) = HK1 (1, K2 ) − (1, K1 + K2 ) = HK1 HK2 − HK1 +K2

(8)

2 − HK1 +K1 2(2, K1) = HK1 (1, K1 ) − (1, K1 + K1 ) = HK 1

(9)

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R. Jakimczuk

Example 1.4 a) Given the indeterminates x1 , x2 , . . . , xn consider the polynomial (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) = H2 H1 − H3 (1, 4 : 1, 1) = H4 H1 − H5 (1, 3 : 1, 2) = H3 H2 − H5 Substituting these formulas into (10) we obtain the polynomial (5) that correspond to (2, 2 : 1, 1) (2, 2 : 1, 1) =

P (H1 , H2 , H3, H4 , H5 ) H 2 H1 − 2H3 H2 − H4 H1 + 2H5 = 2 (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) Formula (8) gives (1, 2 : 1, 1) = H2 H1 − H3 (1, 5 : 1, 1) = H5 H1 − H6 (1, 4 : 1, 2) = H4 H2 − H6 (1, 8 : 1, 1) = H8 H1 − H9 (1, 7 : 1, 2) = H7 H2 − H9

(13)

Symmetric polynomials and number theory

1307

(1, 3 : 1, 1) = H3 H1 − H4 (1, 6 : 1, 3) = H6 H3 − H9 (1, 3 : 1, 2) = H3 H2 − H5

(14)

Substituting (14) into (13) we obtain (1, 3 : 1, 2 : 1, 1) = H3 H2 H1 − H32 − H5 H1 − H4 H2 + 2H6 (1, 6 : 1, 2 : 1, 1) = H6 H2 H1 − H6 H3 − H8 H1 − H7 H2 + 2H9 (1, 5 : 1, 3 : 1, 1) = H5 H3 H1 − H5 H4 − H8 H1 − H6 H3 + 2H9 (1, 4 : 1, 3 : 1, 2) = H4 H3 H2 − H5 H4 − H7 H2 − H6 H3 + 2H9

(15)

Finally, substituting (15) into (12) we obtain the polynomial (5) that correspond to (2, 3 : 1, 2 : 1, 1) 

(2, 3 : 1, 2 : 1, 1) = H32 H2 H1 − H33 − 2H5 H3 H1 − 2H4 H3 H2 − H6 H2 H1 + 5H6 H3 + 2H8 H1 + 2H7 H2 + 2H5 H4 − 6H9 ) /2

(16)

c) We have (formula (6)) 2(2, 1) = H1 (1, 1) − (1, 2) = H1 H1 − H2

(17)

We have (formulas (6) and (7)) 3(3, 1) = H1 (2, 1) − (1, 2 : 1, 1) (1, 2 : 1, 1) = H2 H1 − H3 Therefore 3(3, 1) = H1 (2, 1) − H2 H1 + 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) = H3 H1 − H4 Therefore 4(4, 1) = H1 (3, 1) − H2 (2, 1) + H3 H1 − H4

(19)

In this way we obtain the general recursive formula k(k, 1) = H1 (k − 1, 1) − H2 (k − 2, 1) + . . . + (−1)k Hk−1 H1 + (−1)k+1 Hk (20)

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R. Jakimczuk

where k ≥ 2. (17) gives

H12 − H2 2!

(21)

H13 − 3H2 H1 + 2H3 3!

(22)

H14 − 6H2 H12 + 8H3 H1 + 3H22 − 6H4 4!

(23)

(2, 1) = (18) and (21) give (3, 1) = (19), (21) and (22) give (4, 1) =

(20) with k = 5, (23), (22) and (21) give H15 − 10H2 H13 + 20H3 H12 + 15H22 H1 − 30H4H1 − 20H3 H2 + 24H5 (5, 1) = 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. Property 1. In (5) we have A = L1 !L2 ! . . . Lm !. For example: In (11) we have 2=2! 1!. In (16) we have 2=2!1!1!. In (21) we have 2!. In (22) we have 3!. In (23) we have 4!. In (24) we have 5!. Proof. The property is true if L = 2 since (see (8) and (9)) (1, K1 : 1, K2 ) = HK1 HK2 − HK1 +K2 =

HK1 HK2 − HK1 +K2 1!1!

(25)

2 − HK1 +K1 H 2 − HK1 +K1 HK 1 = K1 (26) 2 2! Suppose that the property is true for L − 1 ≥ 2, we shall prove that the property is true for L. From the inductive hypothesis we can write (see (6), (5) and remarks 1 and 2)

(2, K1 ) =

(L1 − 1, K1 : L2 , K2 : . . . : Lm , Km ) =

P0 (Hi1 , Hi2 , . . . , Hik0 ) (L1 − 1)!L2 ! . . . Lm !

(1, K1 + K1 : L1 − 2, K1 : L2 , K2 : . . . : Lm , Km ) =

(27)

P1 (Hi1 , Hi2 , . . . , Hik1 ) (28) 1!(L1 − 2)!L2 ! . . . Lm !

(1, K1 + K2 : L1 − 1, K1 : L2 − 1, K2 : . . . : Lm , Km ) P2 (Hi1 , Hi2 , . . . , Hik2 ) = 1!(L1 − 1)!(L2 − 1)! . . . Lm !

(29)

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Symmetric polynomials and number theory

.. . (1, K1 + Km : L1 − 1, K1 : L2 , K2 : . . . : Lm − 1, Km ) Pm (Hi1 , Hi2 , . . . , Hikm ) = 1!(L1 − 1)!L2 ! . . . (Lm − 1)!

(30)

Consequently (6) becomes (see (5)) HK1 P0 (Hi1 , Hi2 , . . . , Hik0 ) L1 !L2 ! . . . Lm ! (L1 − 1)P1 (Hi1 , Hi2 , . . . , Hik1 ) L2 P2 (Hi1 , Hi2 , . . . , Hik2 ) − − L1 !L2 ! . . . Lm ! L1 !L2 ! . . . Lm !  Lm Pm (Hi1 , Hi2 , . . . , Hikm ) 1 HK1 P0 (Hi1 , Hi2 , . . . , Hik0 ) − ...− = L1 !L2 ! . . . Lm ! L1 !L2 ! . . . Lm ! − (L1 − 1)P1 (Hi1 , Hi2 , . . . , Hik1 ) − L2 P2 (Hi1 , Hi2 , . . . , Hik2 ) − . . . (L1 , K1 : L2 , K2 : . . . : Lm , Km ) =



− Lm Pm (Hi1 , Hi2 , . . . , Hikm ) =

P (Hi1 , Hi2 , . . . , Hik ) L1 !L2 ! . . . Lm !

(31)

as we desired. Analogously, from the inductive hypothesis we can write (see (7), (5) and remarks 1 and 2) (L2 , K2 : . . . : Lm , Km ) =

P0 (Hi1 , Hi2 , . . . , Hik0 ) L2 ! . . . Lm !

(1, K1 + K2 : L2 − 1, K2 : . . . : Lm , Km ) =

P2 (Hi1 , Hi2 , . . . , Hik2 ) 1!(L2 − 1)! . . . Lm !

.. . (1, K1 + Km : L2 , K2 : . . . : Lm − 1, Km ) =

Pm (Hi1 , Hi2 , . . . , Hikm ) 1!L2 ! . . . (Lm − 1)!

Consequently (7) becomes (see (5)) HK1 P0 (Hi1 , Hi2 , . . . , Hik0 ) 1!L2 ! . . . Lm ! L2 P2 (Hi1 , Hi2 , . . . , Hik2 ) Lm Pm (Hi1 , Hi2 , . . . , Hikm ) −...− − 1!L2 ! . . . Lm ! 1!L2 ! . . . Lm !  1 = HK1 P0 (Hi1 , Hi2 , . . . , Hik0 ) − L2 P2 (Hi1 , Hi2 , . . . , Hik2 ) 1!L2 ! . . . Lm !  P (Hi1 , Hi2 , . . . , Hik ) (32) − . . . − Lm Pm (Hi1 , Hi2 , . . . , Hikm ) = 1!L2 ! . . . Lm ! (1, K1 : L2 , K2 : . . . : Lm , Km ) =

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R. Jakimczuk

as we desired. The property is thus proved. Property 2. The polynomial P (Hi1 , Hi2 , . . . , Hik ) in (5) has an unique monoL1 L2 Lm mial with L factors Hi . This monomial is HK HK . . . HK and its coefficient m 1 2 is 1. The number of factors in the others monomials is less than L. For example: In (11) this unique monomial is H22 H1 . In (16) this unique monomial is H32 H2 H1 . In (24) this unique monomial is H15 . Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2, we shall prove that the property is also true for L. From the inductive hypothesis the polynomial (see (27)) P0 (Hi1 , Hi2 , . . . , Hik0 ) L1 −1 L2 Lm has an unique monomial with L − 1 factors Hi , namely HK HK 2 . . . HK . m 1 This monomial has coefficient 1 and the number of factors Hi in the rest of the monomials is less than L − 1. Consequently the polynomial (see (31)) HK1 P0 (Hi1 , Hi2 , . . . , Hik0 ) will have an unique monomial with L factors Hi , L1 L2 Lm namely HK HK . . . HK . This monomial will have coefficient 1 and the numm 1 2 ber of factors Hi in the rest of the monomials will be less than L. The rest of the polynomials in (31) (inductive hypothesis) have in their monomials a number of factors Hi less than or equal to L − 1. As we desired. The proof using (32) is the same. The property is proved. Property 3. The sum of the coefficients in the polynomial P (Hi1 , Hi2 , . . . , Hik ) is zero. For example in (23) we have 1 − 6 + 8 + 3 − 6 = 0. Proof. The property is true if L = 2 (see (25) and (26)). The proof is now an immediate consequence of the inductive hypotheis and (31) (or (32)). The property is proved. Property 4. If L is even: The monomials, in the polynomial P (Hi1 , Hi2 , . . . , Hik ), that have an odd number of factors Hi have negative coefficient. The monomials, in the polynomial P (Hi1 , Hi2 , . . . , Hik ), that have an even number of factors Hi have positive coefficient. On the other hand, if L is odd: The monomials, in the polynomial P (Hi1 , Hi2 , . . . , Hik ), that have an odd number of factors Hi have positive coefficient. The monomials, in the polynomial P (Hi1 , Hi2 , . . . , Hik ), that have an even

Symmetric polynomials and number theory

1311

number of factors Hi have negative coefficient. For example see (11), (16), (21), (22), (23) and (24). Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 even. Then, an immediate consequence of the inductive hypothesis and (31) (or (32)) is that the property is also true for L odd. Suppose that the property is true for L − 1 odd. Then, an immediate consequence of the inductive hypothesis and (31) (or (32)) is that the property is also true for L even. The property is proved. Property 5. The sum of the absolute values of the coefficients in the polynomial P (Hi1 , Hi2 , . . . , Hik ) is L!. For example: In (16) where L = 4 we have 1+1+2+2+1+5+2+2+2+6 = 4!. In (22) where L = 3 we have 1 + 3 + 2 = 3!. Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2. Then, an immediate consequence of the property 4 and the inductive hypothesis is that the sum of the absolute values of the coefficients in the polynomial P (Hi1 , Hi2 , . . . , Hik ) will be (see (31) and (2)) (L − 1)! + (L1 − 1)(L − 1)! + L2 (L − 1)! + . . . + Lm (L − 1)! = (1 + L1 − 1 + L2 + . . . + Lm )(L − 1)! = L! as desired. The proof using (32) is the same. The property is proved. Property 6. In each monomial of the polynomial P (Hi1 , Hi2 , . . . , Hik ) the sum of the subscripts i in the factors Hi is t (see (3)). Therefore each monomial can be associated to a partition of t. For example in (16), where t = 9 , the sum of subscripts in each monomial is 9. The monomial H32 H2 H1 is associated to the partition 3 + 3 + 2 + 1 = 9. The monomial H33 is associated to the partition 3 + 3 + 3 = 9. The monomial H5 H3 H1 is associated to the partition 5 + 3 + 1 = 9. The monomial H7 H2 is associated to the partition 7 + 2 = 9. The monomial H9 is associated to the partition 9 = 9. Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2.

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Then, an immediate consequence of the inductive hypothesis, remark 2 and (31) (or (32)) is that the property is also true for L. The property is proved. Property 7. Consider the polynomial (5), that is P (Hi1 , Hi2 , . . . , Hik ), and consider the set of L numbers ⎧ ⎪ ⎨

L1





L2







Lm



⎫ ⎬ ⎪

K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km ⎪ ⎪ ⎩ ⎭ whose sum is t (see (3)). Now, consider a partition of this set. If we sum the numbers in each subset of the partition we obtain a partition of t. All monomials in the polynomial (5) are obtained from all these possible partitions of t. Example 1.5 Consider the polynomial (2, 2 : 1, 1) in example 1.4 a). The set of L = 3 elements is {2, 2, 1} and t = 5. All possibles partitions of this set are: {2, 2, 1} {2, 2} ∪ {1} {2, 1} ∪ {2} {2} ∪ {2} ∪ {1} Consequently we obtain the following partitions of 5. 5=5 4+1=5 3+2=5 2+2+1=5 All possible monomials in the polynomial P (Hi1 , Hi2 , . . . , Hik ) will be H5 H4 H1 H3 H2 H22 H1 See (11).

1313

Symmetric polynomials and number theory

In the case of the polynomials (k, 1) of example 1.4 c) where L = t = k the set is ⎧ ⎫ ⎪ ⎨ ⎪ ⎩

k



⎬ ⎪

1, . . . , 1,

⎪ ⎭

Consequently all possible partitions of this set give us all possible partitions of t. See (21), (22), (23) and (24). Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2. Then, an immediate consequence of the inductive hypothesis , (27), (28), (29), . . . (30) and (31) is that the property is also true for L. Note that to the polynomial (27), that is P0 (Hi1 , Hi2 , . . . , Hik0 ), correspond the set of L − 1 numbers ⎧ ⎪ ⎨

L1 −1





L2







⎫ ⎬ ⎪

Lm



K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km

⎪ ⎩

⎪ ⎭

whose sum is t − K1 Note that to the polynomial (28), that is P1 (Hi1 , Hi2 , . . . , Hik1 ) , correspond the set of L − 1 numbers ⎧ ⎪ ⎨ ⎪ ⎩





L1 −2



L2









Lm



⎫ ⎬ ⎪

K1 + K1 , K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km ⎪ ⎭

whose sum is t Note that to the polynomial (29), that is P2 (Hi1 , Hi2 , . . . , Hik2 ) , correspond the set of L − 1 numbers ⎧ ⎪ ⎨ ⎪ ⎩





L1 −1



L2 −1









Lm



⎫ ⎬ ⎪

K1 + K2 , K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km ⎪ ⎭

whose sum is t

.. .

Note that to the polynomial (30), that is Pm (Hi1 , Hi2 , . . . , Hikm ) , correspond the set of L − 1 numbers ⎧ ⎪ ⎨ ⎪ ⎩





L1 −1





L2







Lm −1



⎫ ⎬ ⎪

K1 + Km , K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km

whose sum is t

⎪ ⎭

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Finally, note that to the polynomial P (Hi1 , Hi2 , . . . , Hik ) (see (5) and (31)) correspond the set of L numbers ⎧ ⎪ ⎨

L1





L2







Lm



⎫ ⎬ ⎪

K1 , . . . , K1 , K2 , . . . , K2 , . . . , Km , . . . , Km ⎪ ⎪ ⎩ ⎭ whose sum is t (see the enunciate of property 7). The proof using (32) is the same. The property is proved. Property 8. The polynomial (5) is the zero polynomial if the number of indeterminates xi in the Hi is less than L. For example, consider the polynomial (11) where L = 3. Consequently: (x21 )2 x1 − 2x31 x21 − x41 x1 + 2x51 =0 2 (x21 + x22 )2 (x1 + x2 ) − 2(x31 + x32 )(x21 + x22 ) − (x41 + x42 )(x1 + x2 ) + 2(x51 + x52 ) =0 2

Proof. This property also can be proved without difficulty from (6) or (7). Note that if the number of indeterminates xi is L − 1 then the right side in (6) or (7) is the zero polynomial. In the following section we shall need these properties.

2

Applications to the Number Theory

Now, we shall denote the polynomial (L1 , K1 : L2 , K2 : . . . : Lm , Km ) in the form (L1 , K1 : L2 , K2 : . . . : Lm , Km )(x1 , . . . , xn ) Consider a set of different positive integers {a1 , a2 , . . . , an }. If we put xi = ai then (see definition 1.1) (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) is the sum of all possible products which have t factors ai , L different factors ai and furthermore L1 different factors ai have exponent K1 , L2 different factors ai have exponent K2 , . . ., Lm different factors ai have exponent Km .

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Symmetric polynomials and number theory

Given a strictly increasing sequence an of positive integers we are interested in this section in the asymptotic study ( or exact study, if is possible) of the sequence, (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) Our objective is to obtain in short proofs general formulas applicable to very general sequences an . The properties obtained in the former section will be very useful in the proofs of our main theorems in this section. We shall need the following well known lemma (see [4], page 332). 



and ∞ Lemma 2.1 Let ∞ i=1 ai  i=1 bi be two series of positive terms such ai ∞ that limi→∞ bi = 1. Then if i=1 bi is divergent, the following limit holds n

ai =1 i=1 bi

lim i=1 n n→∞

Now, we shall establish a general theorem. Particular cases of this theorem are well known (see example 2.5). Theorem 2.2 Let f (x) be a function with continuous derivative in a certain interval [a, ∞) such that f (x) > 0, f  (x) > 0, limx→∞ f (x) = ∞ and xf  (x) =0 x→∞ f (x)

(33)

lim

Let an (n ≥ 1) be a strictly increasing sequence of positive integers such that an ∼ ns f (n)

(s ≥ 1)

(34)

Then the following asymptotic formula holds n  α i=1

ai ∼

n nsα+1 f (n)α ∼ aα sα + 1 sα + 1 n

(α > 0)

(35)

Proof. Let us consider the two series 1 + 2 + . . . + (n − 1) +

n 

(is f (i))α

i=n

and

n  i=1

aαi

Where n is a positive integer in the interval [a, ∞). Since xs f (x) is increasing we have n  i=n

s

α

(i f (i)) =

 n n

xsα f (x)α dx + O (nsα f (n)α )

(36)

1316

R. Jakimczuk

On the other hand (L’Hospital’s rule) lim x→∞ Consequently

 n n

 x sα t f (t)α n

dt

=1

xsα+1 f (x)α sα+1

xsα f (x)α dx ∼

nsα+1 f (n)α sα + 1

(37)

(36) and (37) give 1 + 2 + . . . + (n − 1) +

n 

(is f (i))α ∼

i=n

nsα+1 f (n)α n ∼ aα sα + 1 sα + 1 n

(38)

Finally, from (38) and lemma 2.1 we obtain (35). The theorem is thus proved. Remark 3. Note that if an is a strictly increasing sequence of positive integers such that an ∼ ns f (n) we have s ≥ 1. Since (see (33)) d dx Therefore

f (x) xβ



f (x) xβ



f (x) = β+1 x





xf  (x) −β < 0 f (x)

(β > 0)

is decreasing for all β > 0. Consequently

f (x) f (x) 1 = lim β λ−β = 0 λ x→∞ x x→∞ x x lim

(λ > 0)

(0 < β < λ)

(39)

Now, if s < 1 equations (39) and (34) are an evident contradiction. Remark 4. Note that there ever exists a strictly increasing sequence an that satisfies (34), for example an = [ns f (n)], since ((n+1)s f (n+1)−ns f (n)) → ∞. The following general theorem is our first main result. Theorem 2.3 If an ∼ ns f (n) (s ≥ 1) , where f (x) satisfies the conditions of theorem 2.2, the following asymptotic formula holds (see (2) and (3)) (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) 1 1 1 1 ∼ . . . nL atn L L L1 !L2 ! . . . Lm ! (sK1 + 1) 1 (sK2 + 1) 2 (sKm + 1)Lm 1 1 1 1 ∼ ... nst+L f (n)t L L 1 2 L1 !L2 ! . . . Lm ! (sK1 + 1) (sK2 + 1) (sKm + 1)Lm Proof. If r is a positive integer then (see (35)) ar1 + ar2 + . . . + arn ∼

n ar sr + 1 n

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Symmetric polynomials and number theory

That is ar1 + ar2 + . . . + arn =

n arn hr (n) sr + 1

Where hr (n) → 1. Consequently if k is a positive integer we have (ar1 + ar2 + . . . + arn )k =

nk ark hk (n) (sr + 1)k n r

(40)

Let us consider a monomial Hcb11 Hcb22 . . . Hcbkk

(41)

in the polynomial P (Hi1 , Hi2 , . . . , Hik ) (see (5)). (40) gives 

Hcb11 Hcb22 . . . Hcbkk





nb1 nb2 c1 b1 b1 = a h (n) ac2 b2 hbc22 (n) . . . c1 (sc1 + 1)b1 n (sc2 + 1)b2 n   nbk ck bk bk a hck (n) (sck + 1)bk n nb1 +b2 +...+bk = at (sc1 + 1)b1 (sc2 + 1)b2 . . . (sck + 1)bk n hbc11 (n) hbc22 (n) . . . hbckk (n) (42)

That is Hcb11 Hcb22 . . . Hcbkk ∼

nb1 +b2 +...+bk at (sc1 + 1)b1 (sc2 + 1)b2 . . . (sck + 1)bk n

Note that we have used the property 6 (c1 b1 + c2 b2 + . . . + ck bk = t). Now, b1 + b2 + . . . + bk is the number of factors in the monomial (41). Consequently by the property 2 and the property 1 we have (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) 1 1 1 1 ∼ . . . nL atn L L L1 !L2 ! . . . Lm ! (sK1 + 1) 1 (sK2 + 1) 2 (sKm + 1)Lm The theorem is proved. Corollary 2.4 The following asymptotic formula holds (k, 1)(a1 , . . . , an ) ∼

nk akn (a1 + a2 + . . . + an )k n(s+1)k f (n)k ∼ ∼ k! k!(s + 1)k k!(s + 1)k

1318

R. Jakimczuk

Example 2.5 If ai = pi where pi is the i-th prime number we have (Prime Number Theorem) pi ∼ i log i. Consequently theorem 2.2 is applicable and we obtain n nα+1 logα n pαn ∼ pα1 + pα2 + . . . + pαn ∼ α+1 α+1 This formula was first obtained in [5]. There exist more precise formulas when α is a positive integer (see [2]). Therefore, we have (L1 , K1 : L2 , K2 : . . . : Lm , Km )(p1 , . . . , pn ) 1 1 1 1 . . . nL ptn ∼ L1 !L2 ! . . . Lm ! (K1 + 1)L1 (K2 + 1)L2 (Km + 1)Lm 1 1 1 1 ∼ . . . nt+L logt n L L L1 !L2 ! . . . Lm ! (K1 + 1) 1 (K2 + 1) 2 (Km + 1)Lm If ai = ci,k where ci,k is the i-th number with k prime factors in its prime (k−1)!i log i . Consequently theorem 2.2 is factorization we have (see [1]) ci,k ∼ (log log i)k−1 applicable and we obtain cα1,k + cα2,k + . . . + cαn,k

n nα+1 ∼ cαn,k ∼ α+1 α+1



(k − 1)! log n (log log n)k−1



This formula was also obtained in [1]. Therefore, we have (L1 , K1 : L2 , K2 : . . . : Lm , Km )(c1,k , . . . , cn,k ) 1 1 1 1 ... nL ctn,k ∼ L L 1 2 L1 !L2 ! . . . Lm ! (K1 + 1) (K2 + 1) (Km + 1)Lm 1 1 1 1 ∼ ... nt+L L L 1 2 L1 !L2 ! . . . Lm ! (K1 + 1) (K2 + 1) (Km + 1)Lm 

(k − 1)! log n (log log n)k−1

t

We shall now establish a theorem similar to theorem 2.2. Theorem 2.6 If

an ∼ Ans

(43)

where A > 0 and s ≥ 1, (35) holds. Proof. We have n 

(Ais )α = Aα

i=1



n 

isα = Aα

i=1 α sα+1

 n 0

xsα dx + O(nsα ) =

n A n ∼ aα sα + 1 sα + 1 n

Aα nsα+1 + O(nsα ) sα + 1 (44)

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Symmetric polynomials and number theory

The theorem is a direct consequence of (43), (44) and lemma 2.1. The proof is complete. The following theorem is our second main result. Theorem 2.7 If an ∼ Ans , where A > 0 and s ≥ 1, the following asymptotic formula holds (see (2) and (3)) (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) 1 1 1 1 ∼ ... nL atn L L 1 2 L1 !L2 ! . . . Lm ! (sK1 + 1) (sK2 + 1) (sKm + 1)Lm 1 1 1 1 ∼ ... At nst+L L L L m 1 2 L1 !L2 ! . . . Lm ! (sK1 + 1) (sK2 + 1) (sKm + 1) Proof. The proof is the same as theorem 2.3 since (35) holds (theorem 2.6). Corollary 2.8 The following asymptotic formula holds (k, 1)(a1 , . . . , an ) ∼

nk akn (a1 + a2 + . . . + an )k n(s+1)k Ak ∼ ∼ k! k!(s + 1)k k!(s + 1)k

In the following theorem (our last main result) we study polynomial sequences an . In this case we shall see there exist exact formulas. Theorem 2.9 Let P (n) be a polynomial of integer coefficients, degree s ≥ 1 and leading coefficient A > 0 such that 0 < P (1) < P (2) < P (3) < . . . Let us consider the sequence an = P (n) (n ≥ 1). Then (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an )

(45)

is a polynomial in n of rational coefficients, degree st + L (see (2) and (3)) and leading coefficient 1 1 1 1 ... At L L 1 2 L1 !L2 ! . . . Lm ! (sK1 + 1) (sK2 + 1) (sKm + 1)Lm

(46)

This polynomial has the factors n, n − 1, . . . , n − (L − 1). If P (n) has not a constant term then this polynomial has also the factor n + 1. Proof. It is well known (see [3], page 9, for an elementary proof) 1r + 2r + . . . + nr = Qr (n)

(r ≥ 1)

1320

R. Jakimczuk

where Qr (n) is a polynomial in n of rational coefficients, degree r + 1 and leading coefficient 1/(r + 1). Furthermore, these polynomials have the factors n + 1 and n. Since they have the roots 0 and −1. For example 1 1 Q1 (n) = 1 + 2 + . . . + n = n2 + n 2 2 1 1 1 Q2 (n) = 12 + 22 + . . . + n2 = n3 + n2 + n 3 2 6 1 1 1 Q3 (n) = 13 + 23 + . . . + n3 = n4 + n3 + n2 4 2 4 1 1 1 1 Q4 (n) = 14 + 24 + . . . + n4 = n5 + n4 + n3 − n 5 2 3 30 1 1 5 1 Q5 (n) = 15 + 25 + . . . + n5 = n6 + n5 + n4 − n2 6 2 12 12 Consequently P (1)r + P (2)r + . . . + P (n)r will be also a polynomial in n of rational coefficients which we shall denote Tr (n). Note that Tr (n) has the factor n and if P (n) has not a constant coefficient then Tr (n) has also the factor n + 1. (5) gives (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) =

P (Ti1 (n), Ti2 (n), . . . , Tik (n)) A

Therefore (45) is a polynomial in n of rational coefficients. Now, an = P (n) ∼ Ans . Consequently theorem 2.7 is applicable and we obtain (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an ) 1 1 1 1 ∼ ... At nst+L L L 1 2 L1 !L2 ! . . . Lm ! (sK1 + 1) (sK2 + 1) (sKm + 1)Lm That is, (45) has degree st + L and leading coefficient (46). Finally, (45) has the factor n since the polynomials Tr (n) have this factor (see above). On the other hand, property 8 imply that (45) has also the factors n − 1, . . . , n − (L − 1). If P (n) has not a constant coefficient then (45) has also the factor n + 1 since in this case the polynomials Tr (n) have this factor (see above). The theorem is proved. Example 2.10 a) Consider the sequence an = n.

Symmetric polynomials and number theory

1321

Then the polynomial in n (L1 , K1 : L2 , K2 : . . . : Lm , Km )(1, . . . , n) has degree t + L (see (2) and (3)) and leading coefficient 1 1 1 1 . . . L L L1 !L2 ! . . . Lm ! (K1 + 1) 1 (K2 + 1) 2 (Km + 1)Lm since s = 1 and A = 1. Besides, this polynomial has the factors n + 1, n, n − 1, . . . , n − (L − 1). Note that if n = p − 1 , where p is prime, then (L1 , K1 : L2 , K2 : . . . : Lm , Km )(1, . . . , p − 1) is multiple of p except by a finite number of primes. Examples: We have (see 11) 20n8 − 16n7 − 91n6 + 35n5 + 124n4 − 19n3 − 54n2 (2, 2 : 1, 1)(1, . . . , n) = 720 (n + 1)n(n − 1)(n − 2)(20n4 + 24n3 − 23n2 − 27n) = 720 We have (see (21), (22) and (23)) (2, 1)(1, . . . , n) = (3, 1)(1, . . . , n) =

(n + 1)n(n − 1)(3n + 2) 3n4 + 2n3 − 3n2 − 2n = 24 24

(n + 1)n(n − 1)(n − 2)(n2 + n) n6 − n5 − 3n4 + n3 + 2n2 = 48 48

15n8 − 60n7 − 10n6 + 192n5 − 25n4 − 180n3 + 20n2 + 48n 5760 (n + 1)n(n − 1)(n − 2)(n − 3)(15n3 + 15n2 − 10n − 8) = 5760

(4, 1)(1, . . . , n) =

The following asymptotic formula holds (k, 1)(1, . . . , n) ∼

n2k (1 + 2 + . . . + n)k ∼ k! k!2k

b) Let us consider the linear sequence an = An−ri where ri ∈ {0, 1, . . . , A− 1}. Then the polynomial in n (L1 , K1 : L2 , K2 : . . . : Lm , Km )(a1 , . . . , an )

1322

R. Jakimczuk

has degree t + L (see (2) and (3)) and leading coefficient 1 1 1 1 . . . At L1 !L2 ! . . . Lm ! (K1 + 1)L1 (K2 + 1)L2 (Km + 1)Lm Besides, this polynomial has the factors n, n−1, . . . , n−(L−1). The following asymptotic formula holds nk akn (a1 + a2 + . . . + an )k n2k Ak ∼ (k, 1)(a1 , . . . , an ) ∼ ∼ k! k!2k k!2k Example: If an = 2n − 1 then (see (21)) (2, 1)(a1 , . . . , an ) =

3n4 − 4n3 + n n(n − 1)(3n2 − n − 1) = 6 6

References [1] R. Jakimczuk, A note on sums of powers which have a fixed number of prime factors, Journal of Inequalities in Pure and Applied Mathematics, (2005), volume 6, issue 2, article 31. [2] R. Jakimczuk, Desigualdades y f´ormulas asint´oticas para sumas de potencias de primos, Bolet´in de la Sociedad Matem´ atica Mexicana, (2005), volumen 11, n´ umero 1, 5-10. [3] W. J. LeVeque, Topics in Number Theory, volume I, Addison-Wesley, First Edition, 1958. [4] J. Rey Pastor, P. Pi Calleja, C. Trejo, An´ alisis Matem´ atico, volumen I, Editorial Kapeluz, Octava Edici´ on, 1969. [5] T. S´alat and S. Zn´am, On the sums of prime powers, Acta Fac. Rer. Nat. Univ. Com. Math., 21 (1968), 21 - 25. [6] B. L. Van Der Waerden, Modern Algebra, volume I, Frederick Ungar Publishing Co., New York, 1948. Received: November, 2009