International Mathematical Forum, 5, 2010, no. 27, 1303 - 1322

On Symmetric 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 xk1xk2 ...xkh 1 2 h

in n indeterminates x1,x2,...,xn may be expressed as a 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 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 of the indeterminates is called a . 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 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.

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)) H H − H K1 K2 K1+K2 (1,K :1,K )=HK HK − HK K = (25) 1 2 1 2 1+ 2 1!1! 2 2 H − HK K H − HK K (2,K )= K1 1+ 1 = K1 1+ 1 (26) 1 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)

P Hi ,Hi ,...,Hi 0( 1 2 k0 ) (L1 − 1,K1 : L2,K2 : ...: Lm,Km)= (27) (L1 − 1)!L2! ...Lm!

P Hi ,Hi ,...,Hi 1( 1 2 k1 ) (1,K1 +K1 : L1 −2,K1 : L2,K2 : ...: Lm,Km)= (28) 1!(L1 − 2)!L2! ...Lm!

(1,K1 + K2 : L1 − 1,K1 : L2 − 1,K2 : ...: Lm,Km) P H ,H ,...,H 2( i1 i2 ik ) = 2 (29) 1!(L1 − 1)!(L2 − 1)! ...Lm! Symmetric polynomials and number theory 1309

. .

(1,K1 + Km : L1 − 1,K1 : L2,K2 : ...: Lm − 1,Km) Pm(Hi ,Hi ,...,Hi ) = 1 2 km (30) 1!(L1 − 1)!L2! ...(Lm − 1)!

Consequently (6) becomes (see (5))

HK P Hi ,Hi ,...,Hi 1 0( 1 2 k0 ) (L1,K1 : L2,K2 : ...: Lm,Km)= L1!L2! ...Lm! (L − 1)P (Hi ,Hi ,...,Hi ) L P (Hi ,Hi ,...,Hi ) − 1 1 1 2 k1 − 2 2 1 2 k2 L1!L2! ...Lm! L1!L2! ...Lm! L P H ,H ,...,H m m( i1 i2 ikm ) 1 − ...− HK P Hi ,Hi ,...,Hi = 1 0( 1 2 k0 ) L1!L2! ...Lm! L1!L2! ...Lm! − L − P Hi ,Hi ,...,Hi − L P Hi ,Hi ,...,Hi − ... ( 1 1) 1( 1 2 k1 ) 2 2( 1 2 k2 ) P (Hi ,Hi ,...,Hi ) − L P H ,H ,...,H 1 2 k m m( i1 i2 ikm ) = (31) L1!L2! ...Lm! as we desired. Analogously, from the inductive hypothesis we can write (see (7), (5) and remarks 1 and 2)

P Hi ,Hi ,...,Hi 0( 1 2 k0 ) (L2,K2 : ...: Lm,Km)= L2! ...Lm!

P Hi ,Hi ,...,Hi 2( 1 2 k2 ) (1,K1 + K2 : L2 − 1,K2 : ...: Lm,Km)= 1!(L2 − 1)! ...Lm! . . P H ,H ,...,H m( i1 i2 ikm ) (1,K1 + Km : L2,K2 : ...: Lm − 1,Km)= 1!L2! ...(Lm − 1)! Consequently (7) becomes (see (5))

HK P Hi ,Hi ,...,Hi 1 0( 1 2 k0 ) (1,K1 : L2,K2 : ...: Lm,Km)= 1!L2! ...Lm! L P H ,H ,...,H ( i1 i2 ik ) LmPm(Hi ,Hi ,...,Hi ) − 2 2 2 − ...− 1 2 km 1!L2! ...Lm! 1!L2! ...Lm! 1 HK P Hi ,Hi ,...,Hi − L P Hi ,Hi ,...,Hi = 1 0( 1 2 k0 ) 2 2( 1 2 k2 ) 1!L2! ...Lm! P (Hi ,Hi ,...,Hi ) − ...− L P H ,H ,...,H 1 2 k m m( i1 i2 ikm ) = (32) 1!L2! ...Lm! 1310 R. Jakimczuk as we desired. The property is thus proved.

Property 2. P H ,H ,...,H The polynomial ( i1 i2 ik ) in (5) has an unique mono- L H HL1 HL2 ...HLm mial with factors i. This is K1 K2 Km and its coefficient is 1. The number of factors in the others monomials is less than L.

H2H For example: In (11) this unique monomial is 2 1. In (16) this unique H2H H H5 monomial is 3 2 1. In (24) this unique monomial is 1 .

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. P Hi ,Hi ,...,Hi From the inductive hypothesis the polynomial (see (27)) 0( 1 2 k0 ) L − H HL1−1HL2 ...HLm has an unique monomial with 1 factors i, namely K1 K2 Km . 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)) HK P Hi ,Hi ,...,Hi L Hi 1 0( 1 2 k0 ) will have an unique monomial with factors , HL1 HL2 ...HLm namely K1 K2 Km . This monomial will have coefficient 1 and the num- 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. P H ,H ,...,H The sum of the coefficients in the polynomial ( i1 i2 ik ) 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: P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an odd number of factors Hi have negative coefficient. P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an even number of factors Hi have positive coefficient. On the other hand, if L is odd: P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an odd number of factors Hi have positive coefficient. P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), 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 poly- P H ,H ,...,H nomial ( i1 i2 ik ) 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 hy- pothesis is that the sum of the absolute values of the coefficients in the poly- P H ,H ,...,H nomial ( i1 i2 ik ) 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. P H ,H ,...,H In each monomial of the polynomial ( i1 i2 ik ) 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 H2H H is 9. The monomial 3 2 1 is associated to the partition 3 + 3 + 2 + 1 = 9. H3 The monomial 3 is associated to the partition 3 + 3 + 3 = 9. The monomial H5H3H1 is associated to the partition 5 + 3 + 1 = 9. The monomial H7H2 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. 1312 R. Jakimczuk

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. P H ,H ,...,H Consider the polynomial (5), that is ( i1 i2 ik ), and consider the set of L numbers ⎧ ⎫ L L L ⎨⎪ 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪ 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 =3elements 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 P H ,H ,...,H All possible monomials in the polynomial ( i1 i2 ik ) will be

H5

H4H1

H3H2 H2H 2 1 See (11). Symmetric polynomials and number theory 1313

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. P Hi ,Hi ,...,Hi Note that to the polynomial (27), that is 0( 1 2 k0 ), correspond the set of L − 1 numbers ⎧ ⎫ L − L L ⎨⎪ 1 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪

whose sum is t − K1 P Hi ,Hi ,...,Hi Note that to the polynomial (28), that is 1( 1 2 k1 ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L L ⎨⎪ 1 2 2 m ⎬⎪ K + K , K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 1 1 2 2 ⎭⎪ whose sum is t P Hi ,Hi ,...,Hi Note that to the polynomial (29), that is 2( 1 2 k2 ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L − L ⎨⎪ 1 1 2 1 m ⎬⎪ K + K , K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 2 1 1 2 2 ⎭⎪ whose sum is t . . P H ,H ,...,H Note that to the polynomial (30), that is m( i1 i2 ikm ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L L − ⎨⎪ 1 1 2 m 1 ⎬⎪ K + Km, K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 1 2 2 ⎭⎪ whose sum is t 1314 R. Jakimczuk

P H ,H ,...,H Finally, note that to the polynomial ( i1 i2 ik) (see (5) and (31)) correspond the set of L numbers ⎧ ⎫ L L L ⎨⎪ 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪ 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:

(x2)2x − 2x3x2 − x4x +2x5 1 1 1 1 1 1 1 =0 2 (x2 + x2)2(x + x ) − 2(x3 + x3)(x2 + x2) − (x4 + x4)(x + x )+2(x5 + x5) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 =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. Symmetric polynomials and number theory 1315

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

n ai lim i=1 =1 n→∞ n b i=1 i 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) lim = 0 (33) x→∞ f(x)

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

s an ∼ n f(n)(s ≥ 1) (34)

Then the following asymptotic formula holds

n nsα+1f n α n aα ∼ ( ) ∼ aα α> i sα sα n ( 0) (35) i=1 +1 +1

Proof. Let us consider the two series n n  s α α 1+2+...+(n − 1) + (i f(i)) and ai i=n i=1

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

n n α (isf(i)) = xsαf(x)α dx + O (nsαf(n)α) (36) n i=n 1316 R. Jakimczuk

On the other hand (L’Hospital’s rule) x sα α n t f(t) dt xlim→∞ xsα+1f(x)α =1 sα+1

Consequently n nsα+1f(n)α xsαf(x)α dx ∼ (37) n sα +1 (36) and (37) give n nsα+1f n α n ... n − isf i α ∼ ( ) ∼ aα 1+2+ +( 1) + ( ( )) sα sα n (38) i=n +1 +1 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 s such that an ∼ n f(n) we have s ≥ 1. Since (see (33)) d f(x) f(x) xf (x) = − β < 0(β>0) dx xβ xβ+1 f(x)

f(x) β> Therefore xβ is decreasing for all 0. Consequently f(x) f(x) 1 lim = lim =0 (λ>0) (0 <β<λ) (39) x→∞ xλ x→∞ xβ xλ−β 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 s s s satisfies (34), for example an =[n f(n)], since ((n+1) f(n+1)−n f(n)) →∞.

The following general theorem is our first main result.

s Theorem 2.3 If an ∼ n 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 at L1 L2 Lm n L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1) 1 1 1 1 ∼ ... nst+L f(n)t L1 L2 Lm L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1) Proof. If r is a positive integer then (see (35)) n ar + ar + ...+ ar ∼ ar 1 2 n sr +1 n Symmetric polynomials and number theory 1317

That is r r r n r a + a + ...+ a = a hr(n) 1 2 n sr +1 n

Where hr(n) → 1. Consequently if k is a positive integer we have

k k n (ar + ar + ...+ ar ) = ark hk(n) (40) 1 2 n (sr +1)k n r

Let us consider a monomial

Hb1 Hb2 ...Hbk c1 c2 ck (41) P H ,H ,...,H in the polynomial ( i1 i2 ik ) (see (5)). (40) gives nb1 nb2 b1 b2 bk c1b1 b1 c2b2 b2 Hc Hc ...Hc = an hc (n) an hc (n) ... 1 2 k (sc +1)b1 1 (sc +1)b2 2 1 2 nbk ackbk hbk n b n ck ( ) (sck +1) k

b1+b2+...+bk n t = an b1 b2 bk (sc1 +1) (sc2 +1) ...(sck +1) hb1 n hb2 n ...hbk n c1 ( ) c2 ( ) ck ( ) (42)

That is

nb1+b2+...+bk b1 b2 bk t Hc Hc ...Hc ∼ an 1 2 k b1 b2 bk (sc1 +1) (sc2 +1) ...(sck +1)

Note that we have used the property 6 (c1b1 + c2b2 + ...+ ckbk = 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 at L1 L2 Lm n L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1)

The theorem is proved.

Corollary 2.4 The following asymptotic formula holds

k k k (s+1)k k (a1 + a2 + ...+ an) n an n f(n) (k, 1)(a ,...,an) ∼ ∼ ∼ 1 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 pα + pα + ...+ pα ∼ pα ∼ 1 2 n α +1 n 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 pt L1 L2 Lm n L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1) 1 1 1 1 ∼ ... nt+L logt n L1 L2 Lm L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1)

If ai = ci,k where ci,k is the i-th number with k prime factors in its prime c ∼ (k−1)!i log i factorization we have (see [1]) i,k (log log i)k−1 . Consequently theorem 2.2 is applicable and we obtain α n nα+1 (k − 1)! log n cα + cα + ...+ cα ∼ cα ∼ 1,k 2,k n,k α +1 n,k α +1 (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 ct L1 L2 Lm n,k L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1) ∼ 1 1 1 ... 1 nt+L L1 L2 Lm L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1) t (k − 1)! log n (log log n)k−1 We shall now establish a theorem similar to theorem 2.2.

Theorem 2.6 If s an ∼ An (43) where A>0 and s ≥ 1, (35) holds. Proof. We have n n n Aαnsα+1 Ais α Aα isα Aα xsα dx O nsα O nsα ( ) = = + ( )= sα + ( ) i=1 i=1 0 +1 Aαnsα+1 n ∼ ∼ aα (44) sα +1 sα +1 n Symmetric polynomials and number theory 1319

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.

s Theorem 2.7 If an ∼ An , where A>0 and s ≥ 1, the following asymp- totic formula holds (see (2) and (3))

(L1,K1 : L2,K2 : ...: Lm,Km)(a1,...,an) ∼ 1 1 1 ... 1 nL at L1 L2 Lm n L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1) ∼ 1 1 1 ... 1 Atnst+L L1 L2 Lm 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 k k (s+1)k k (a1 + a2 + ...+ an) n an n A (k, 1)(a ,...,an) ∼ ∼ ∼ 1 k! k!(s +1)k k!(s +1)k

In the following theorem (our last main result) we study polynomial se- quences 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

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 (46) L1 L2 Lm L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1)

This polynomial has the factors n, n − 1,...,n− (L − 1).IfP (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)

r r r 1 +2 + ...+ n = 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 Q (n)=1+2+...+ n = n2 + n 1 2 2 1 1 1 Q (n)=12 +22 + ...+ n2 = n3 + n2 + n 2 3 2 6 1 1 1 Q (n)=13 +23 + ...+ n3 = n4 + n3 + n2 3 4 2 4 1 1 1 1 Q (n)=14 +24 + ...+ n4 = n5 + n4 + n3 − n 4 5 2 3 30 1 1 5 1 Q (n)=15 +25 + ...+ n5 = n6 + n5 + n4 − n2 5 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 coeffi- cient then Tr(n) has also the factor n +1. (5) gives

P (Ti (n),Ti (n),...,Ti (n)) L ,K L ,K ... L ,K a ,...,a 1 2 k ( 1 1 : 2 2 : : m m)( 1 n)= A

Therefore (45) is a polynomial in n of rational coefficients. s Now, an = P (n) ∼ An . Consequently theorem 2.7 is applicable and we obtain

(L1,K1 : L2,K2 : ...: Lm,Km)(a1,...,an) ∼ 1 1 1 ... 1 Atnst+L L1 L2 Lm L1!L2! ...Lm! (sK1 +1) (sK2 +1) (sKm +1)

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 L1 L2 Lm L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1) since s =1and 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))

3n4 +2n3 − 3n2 − 2n (n +1)n(n − 1)(3n +2) (2, 1)(1,...,n)= = 24 24 n6 − n5 − 3n4 + n3 +2n2 (n +1)n(n − 1)(n − 2)(n2 + n) (3, 1)(1,...,n)= = 48 48

15n8 − 60n7 − 10n6 + 192n5 − 25n4 − 180n3 +20n2 +48n (4, 1)(1,...,n)= 5760 (n +1)n(n − 1)(n − 2)(n − 3)(15n3 +15n2 − 10n − 8) = 5760 The following asymptotic formula holds

(1 + 2 + ...+ n)k n2k (k, 1)(1,...,n) ∼ ∼ 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 L1!L2! ...Lm! (K1 +1) (K2 +1) (Km +1) Besides, this polynomial has the factors n, n−1,...,n−(L−1). The following asymptotic formula holds

k k k 2k k (a1 + a2 + ...+ an) n an n A (k, 1)(a ,...,an) ∼ ∼ ∼ 1 k! k!2k k!2k Example: If an =2n − 1 then (see (21))

3n4 − 4n3 + n n(n − 1)(3n2 − n − 1) (2, 1)(a ,...,an)= = 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 po- tencias 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 Pub- lishing Co., New York, 1948.

Received: November, 2009