A family of sequences generated by eliminating some prime factors from the sequence of the naturals

A family of sequences generated by eliminating some prime factors from

the sequence of the naturals

Zhihe Li

Ningbo Xiaoshi High School

Adviser: Cindy Boyd & David Ochoro

Table of the Paper

Abstract

...03

Section 1

...05

Section 2

...08

Section 3

...15

Section 4

...26

Summary and Prospect.

...40

Reference

...42

ABSTRACT

This paper starts with an interesting property of the sequence of the naturals {A_n}, i.e.

{A_n}: {1, 2,3,...} .Considering the generated sequence { }L_n by eliminating all the factor 2 in A_n if it has, we show that the partial sum of the new sequence { }L_n from 2^n1th term to(2ⁿ1) th term is square number 2^2n2.

This property is further extended to the more general case: _{{ ( )}H n} is the sequence with each term H n( ) generated according to the following transforming rule:

ER({ ,p p_{1 2},...,p_m}): Given m primes p i_i, 1, 2,...,m, for eachi{1, 2,..., }m , if A_n has factorp_i, then substitute A_nwith A_n/ p_i until it does not have factor p_i.

We focus on the partial sum of ^{{ ( )}H n} . In section one, we research on Question 1:

Given one prime ^p, we eliminate all the factor ^pin {A_n} according to the transforming rule ER^{({ })p} . Then the new sequence { }Ln obtained has the property as follows:

1 n 1

n

kp i i kp

L









^{=( 1) 2 2 1}

2 p n

k p ^

 .

The main results in this section are Theorem 1, 2 and the relative corollaries.

In section two, we mainly study Question 2: Given two prime ^{p q,} , we eliminate all factors ^pand ^qin {A_n} according to the transforming rule ER^{({ , })p q} . Then the new sequence { ( )}H n obtained has the property as follows:

1

( )

r s

kp q

i

H i



 ^{＝k pq p2} ₂₍⁽_p²_^r₁₎₍^1)(_q^q2₁₎^s1)+

1 1 1

( ) ( ) ( )

r s

kp kq k

i i i

H i H i H i

  

 

  

（or ( )

r s

r

kp q

i kp

H i



 ^{＝k pq p2} ₂₍⁽_p²_^r₁₎₍^1)(_q^q2₁₎^{s1)+kqs ( )} i k

H i



 ^).

We also focus on the number of valued “1” point of ^{{ ( )}H n} from this section, where the term in {A_n} which becomes the number 1 in ^{{ ( )}H n} is called a valued “1” point of

{ ( )}H n .Let ^{Y n( ){ |i H i( )1,iN i, n}}, we have

| (Y kp q^{r s}) | | (Y kp^r) || (Y kq^s) | | ( ) | Y k rs,

The main results in this section are Theorem 3, 4, 5, 6 and the relative corollaries.

In section three, we extend to the case of given m primes p i_i, 1, 2,...,m, i.e.

Question 3, and obtain similar results in Theorem 7,8,9,10 and the relative corollaries.

In section four, we propose Question 4 and obtain several concise inequalities to estimate the partial sum of the first n terms and the number of valued “1” point in the first n terms. Furthermore, we derive a new proof for the noted fact that the set of primes is infinite.

For given two primes case, we have (Theorem 11,12)

2 2

2 ( ) 2

2( 1)( 1) i x 2( 1)( 1)

pq pq

x x H i x x

p q   _  p q 

 



 

1 log log 1 log log

1 | ( ) | ( 1)( 2)

2 log log 2 log log

x x x x

p q Y x p q

 

    

 

  (wherepq)

For given m primes case, we have (Theorem 13,14)

2 1 2 1

1 1

1 1

2 ( ) 2

2 1 2 1

m m

m m

i i

i x

i i i i

p p

x x H i x x

p p

 



 

   







 

.

1

1

1 log

| ( ) | (log )

! log

m

m m

i i

Y x x O x

m p





 



.

Finally, we point out that the multiplicative functions ^{H n H n( ), ( )} have further research value.

Key words: sequence of the naturals, multiplicative function, partial sum, valued “1”

point.

Section 1

The set of naturals is buried with numerous secrets. Among those, I find an interesting rule. Let {A_n} be the sequence of the naturals i.e.{A_n} {1, 2,3,..., ,...} n .We obtain a new sequence {L_n} by eliminating all the factor 2 for every term of {A_n} according to the following rule:

(1)If An has factor 2,then we substitute An with A_n/ 2 until it does not have factor 2;

(2) If A_n does not have factor 2,then keep it unchanged.

Thus, we obtain a new sequence { }L_n as follows:

1,1,3,1,5,3,7,1,9,5,11,3,13,7,15,1,17,9,19,5,21,11,23,3,25,13,27,7,29,15,31,1,33,17,35,9 ,37,19,39,5,41,… （1）

We can dig out some secrets if we divide it into groups with 1 being the first term of each group as follows:

(1),(1,3),(1,5,3,7),(1,9,5,11,3,13,7,15),(1,17,9,19,5,21,11,23,3,25,13,27,7,29,15,31),(1,3 3,17,35,9,37,19,39,5,41,……

Summing up each group, we derive that the sum of these groups are 1,4,16,64,…

respectively. Thus we have the following conjecture：

Conjecture 1. The sum of the new sequence {L_n} from 2^n1th term to (2ⁿ1) th term is

2 2

2 ^n .

Since {L_n} contains lots of 3,5,…, we also can divide them by setting 3,5…be the first term of each group as follows:

(1,1),(3,1,5),(3,7,1,9,5,11),(3,13,7,15,1,17,9,19,5,21,11,23),(3,25,13,27,7,29,15,31,1,33, 17,35,9,37,19,39,5,41,……

(1,1,3,1),(5,3,7,1,9),(5,11,3,13,7,,15,1,17,9,19),(5,21,11,23,3,25,13,27,7,29,15,31,1,33,1 7,35,9,37,19,39),(5,41,……

We find that if we divide {L_n} by 3, the sums of second, third, fourth,…groups are 9,36,144,… respectively; if we divide it by 5, the sums of second, third, fourth,…groups are 25, 100, 400,… respectively;…

Hence, we put forward the second conjecture.

Conjecture 2. The sum of the new sequence {L_n} from _k₂^n1th term to (k 2ⁿ 1) th term is k²2^2n2,where kN.

The above transformation is eliminating all the factor 2 for every term of {A_n}. But what property does the new sequence have if we eliminate all the factor 3 for every term of {A_n}? Or even more generally, we consider the following question:

Question 1: Let p be a prime. We obtain a new sequence {L_n} by transforming {A_n} according to the rule ER({ })p as follows:

(1)If A_n has factor p, then we substitute A_n with A_n / p until it does not have factor p;

(2) If A_n does not have factor p, then keep it unchanged.

What properties does {L_n} have?

The rule ER({ })p above can also be written as follows:

If A_n  p^rX, wherer0,X 1are integers，_{( , )X p} ₁，then L_n X . For this question, we derive the following theorem.

Theorem 1. Let p be a prime. Let {L_n} be the sequence generated by transforming {A_n} according to the rule ER^{({ })p} of Question 1.Then for ^{kN n, N},the following equality holds:

1 n 1

n

kp i i kp

L









^{=( 1) 2 2 1}

2 p n

k p ^

 （2）

Proof. The sum

1 n 1

n

kp i i kp

L









^{is from kpn1}th term to (kpⁿ1)th term，which means the number

of the terms is kpⁿ 1 kp^n1 1 k p( 1)p^n1.

Now, we claim that the k p( 1)p^n1 terms are different from each other.

Indeed, assume that there are positive integers i and j such that kp^n1  i j kpⁿ1， with Li Lj. Then by the definition of sequence{L_n}, we have that i p^rL_i,j p^sLj

where r and s are nonnegative integers. Under the assumption, we have j i p^{s r} ,s r 1, which contradicts with kp^n1  i j kpⁿ1.

Furthermore, those k p( 1)p^n1 terms are just the positive integers no larger than kpⁿ except those multiples of p. In fact, the set of positive integers which are no larger than kpⁿ contains kp^n1 multiples of p. Thus, there are k p( 1)p^n1 terms left exactly, after these

1

kpn^ multiples of p removed.

Note that the sum of these kp^n1 multiples of p removed is

1

1 kpn

i

p i









，so we have

1 n 1

n

kp i i kp

L









^＝

1 kpn

i

i



 ^{- 1} 1 kpn

i

p i









^＝1



¹



¹



^{1 1}



¹

2 2

n n n n

kp kp p kp ^ kp ^

   

＝¹

 

^{2 1 1}

2 2

n n n

kp  kp kp ^ ＝^{( 1) 2 2 1} 2

p n

k p ^

 □

Theorem 2 is derived from Theorem 1.

Theorem 2. Let p be a prime. Let {L_n} be the sequence generated by transforming {A_n} according to the rule ER({ })p of Question 1. Then for k n, N,the following equality holds:

1 kpn

i i

L



 ^＝ 1

2 2

( 1)

2( 1)

k i i

k p p n

L p





 



（3）

Proof.

1 kpn

i i

L



 ^＝ ₁

1 1 1

j

j

k n kp

i i

i j i kp

L L

   

  



＝

1

2 2 1

1

( 1) 2

k n

i

i j

k j

L p

p







 

 

＝

1

2 2 1

1 1

12

k n

j

j i

i

L k p

  p

 

  

 

 



＝

2 2

2

2 1

1 1 (1 )

12 1

k i i

n

L p p

p k p



   

  

 





＝

1

2 2

( 1)

2( 1)

k i i

k p p n

L p





 



. □ For the case k＝1，the following corollary holds：

Corollary 1. Let p be a prime. Let {L_n} be the sequence generated by transforming {A_n} according to the rule ER^{({ })p} of Question 1.Then for nN,the following equality holds:

（1）

1 n 1

n

p i i p

L









^=(p₂^1)p2n1

（2）

1

1 pn

i i

L





 ^{＝ p p}₂₍⁽ _p²ⁿ_^₁₎¹⁾

For the case p＝2，both conjecture 1 and conjecture 2 are correct.

Section 2

In first section, we study the properties of the new sequence generated by eliminating all the factor p in {A_n}. Naturally, we wish to generalize it for the case of eliminating more primes. In this section, we consider the case of given two primes.

Question 2. Let p and q be primes. We obtain a new sequence^{{ ( )}H n} by transforming {A_n}according to the rule ER^{({ , })p q} as follows:

(1)If A_n has factor p, then we substitute A_n with A_n / p until it does not have factor

p; If A_n has factor q, then we substitute A_n with A_n /q until it does not have factor q;

(2) If An does not have factors p and q, then keep it unchanged.

What properties does ^{{ ( )}H n} have?

The rule ER({ , })p q above can also be written as follows:

If A_n  p q^{r s}X, wherer0,s0,X 1are integers，( ,X pq) 1 ，then H n( )X . For instance, let p=2, q=3. Then the generated sequence ^{{ ( )}H n} is:

1,1,1,1,5,1,7,1,1,5,11,1,13,7,5,1,17,1,19,5,7,11,23,1,25,13,1,7,29,5,31,1,11,17,35,1,37,1 9,13,5,41,7,43,11,…… （4）

This sequence becomes much more complicated. What can be seen is that more number 1 occurs in this new sequence. In the following part, we study the partial sum of this new sequence and the frequency of number 1.

Firstly, the number-theoretic functionH n( ) :NR is a completely multiplicative function (cf. [1], p77-79) where R is the set of real numbers. And the function has following properties.

（1）H p( )H q( )H(1)1;

( ) ,

H p  p when^p is a prime, ^{p p p, q}.

（2）H mn( )H m H n H pn( ) ( ), ( )H n H qn( ), ( )H n( )；

（3）H n( )n,when n does not have factor p or q.

Theorem 3. Let p and q be primes. Let ^{{ ( )}H n} be the sequence generated by transforming {A_n} according to the rule ER^{({ , })p q} of Question 2.Then for kN,the following equality holds:

1

( )

kpq

i

H i



 ^{＝ 2 ( 1)( 1)}

2

k pq p q +

1

( )

kp

i

H i



 ⁺ 1

( )

kq

i

H i



 ^- 1

( )

k

i

H i



 （5）

Note that this equality can be written as the following briefer forms：

( )

kpq

i kp

H i



 ^{＝ 2 ( 1)( 1)}

2

k pq p q + ^{( )}

kq

i k

H i





or ( )

kpq

i kq

H i



 ^{＝ 2 ( 1)( 1)}

2

k pq p q + ^{( )}

kp

i k

H i



 (5) Proof. we divide{ ( ),H i i1, 2,...,kpq}into two groups and sum up for each group.

In the first group, i is neither a multiple of p nor a multiple of q. Then ^{H i( )}＝i. By Principle of cross-classification (cf. [2], p123) ，the sum of this group is

1

1 1 1 1

kpq kq kp k

i i i i

S i p i q i pq i

   







 

 



In the second group, i is a multiple of p or q. By Principle of cross-classification as well, we derive that this group sums is

2

1 1 1

( ) ( ) ( )

kq kp k

i i i

S H pi H qi H pqi

  













^.

Hence

1 2

1

( )

kpq

i

H i S S



 



＝

1 1 1 1

kpq kq kp k

i i i i

i p i q i pq i

   

  

   

⁺

1 1 1

( ) ( ) ( )

kq kp k

i i i

H pi H qi H pqi

  

 

  

＝¹ (1 ) ¹ (1 ) ¹ (1 ) ¹ (1 ) 2kpq kpq 2 p kq kq 2q kp kp 2 pq k k +

1 1 1

( ) ( ) ( )

kq kp k

i i i

H i H i H i

  

 

  

＝ (1 1 1 1 ) 2

kpq kpq kq kp k +

1 1 1

( ) ( ) ( )

kq kp k

i i i

H i H i H i

  

 

  

＝

2

( 1)( 1)

2

k pq p q +

1

( )

kp

i

H i



 ⁺ 1

( )

kq

i

H i



 ^- 1

( )

k

i

H i



 □ Theorem 4. Let p and q be primes. Let^{{ ( )}H n} be the sequence generated by transforming

{A_n}according to the rule ER^{({ , })p q} of Question 2. Then for ^{k r s, , N},the following

equality holds:

1

( )

r s

kp q

i

H i



 ^{＝k pq p2} ₂₍⁽_p²_^r₁₎₍^1)(_q^q2₁₎^s1)+

1 1 1

( ) ( ) ( )

r s

kp kq k

i i i

H i H i H i

  

 

  

. （6）

Similarly，this equality can be written as the following briefer forms：

( )

r s

r

kp q

i kp

H i



 ^{＝ 2 ( 2 1)( 2 1)}

2( 1)( 1)

r s

k pq p q

p q

 

  + ( )

kqs

i k

H i





or ( )

r s

s

kp q

i kq

H i



 ^{＝ 2 ( 2 1)( 2 1)}

2( 1)( 1)

r s

k pq p q

p q

 

  + ( )

kpr

i k

H i



 . (6) Proof. We prove (6) by induction on r and s. It follows from Theorem 3 that (6) holds if r＝1，

s＝1.

Assume that the equality hold for all r≤u，s≤v，then if r＝u+1，s≤v，by assumption we have that

1

1

( )

u s

kp q

i

H i





 ^＝(kp)2₂₍^{pq p}_p⁽_²₁₎₍^{u }_q¹⁾⁽_1)^{q2s1)+( ) ( ) ( )}

1 1 1

( ) ( ) ( )

u s

kp p kp q kp

i i i

H i H i H i

  

 

  

＝

2 2 2

( ) ( 1)( 1)

2( 1)( 1)

u s

kp pq p q

p q

 

  +

1

1 1 1

( ) ( ) ( )

u s

kp kpq kp

i i i

H i H i H i



  

 

  

＝

2 2 2

( ) ( 1)( 1)

2( 1)( 1)

u s

kp pq p q

p q

 

  +

1

1

( )

kpu

i

H i







2 2 2

1 1 1 1

( 1)( 1)

( ) ( ) ( ) ( )

2( 1)( 1)

s kp kqs k kp

i i i i

k pq p q

H i H i H i H i

p q _{   }

   

   















（unfold

1

( )

kpqs

i

H i



 by the induction hypothesis）

＝

2 2

( 1)

2( 1)( 1)

k pq q s

p q



  ×(p²p^2u p²p² 1)+

1

1 1 1

( ) ( ) ( )

u s

kp kq k

i i i

H i H i H i



  

 

  

＝

2 2

( 1)

2( 1)( 1)

k pq q s

p q



  ×(p^2(u1)1)+

1

1 1 1

( ) ( ) ( )

u s

kp kq k

i i i

H i H i H i



  

 

  

^,

then (6) holds for r＝u+1,s≤v.

Similarly one can prove that (6) holds if r≤u+1, s＝v+1.

Hence the conclusion holds. □

In the case k＝1，the following corollary holds：

Corollary2. Let p and q be primes. Let^{{ ( )}H n} be the sequence generated by transforming {A_n}according to the rule ER({ , })p q of Question 2. Then for r s, N ,the following equalities hold:

(1) ( )

pq

i p

H i



 ^{＝ ( 1)( 1)}

2 pq p q

+

1

( )

q

i

H i





（ or ( )

pq

i q

H i



 ^{＝ ( 1)( 1)}

2 pq p q

+

1

( )

p

i

H i



 ^）

(2) ( )

r s

r

p q

i p

H i



 ^{＝ pq p}₂₍⁽ _p^2r_^₁₎₍¹⁾⁽_q^q_^2s₁₎^1)+ 1

( )

qs

i

H i





（ or ( )

r s

s

p q

i q

H i



 ^{＝ ( 2 1)( 2 1)}

2( 1)( 1)

r s

pq p q

p q

 

  +

1

( )

pr

i

H i



 ^）.

Now we work on the times that number 1 occurs in^{{ ( )}H n} . Similar to zero point, we call the term in {A_n} which becomes to the number 1 in the new sequence^{{ ( )}H n} a valued

“1” point of { ( )}H n . We defineY n( ){ |i H i( )1,iN i, n}, and the cardinality of Y(n) is denoted by ^{| ( ) |Y n} .

Our idea is to transform calculating number of valued “1” points to the partial sum of a sequence. Set a number-theoretic function ^{H n( ) :NR} as follows:

( ) ( ) 1

H n H n  ，whenH n( )1；H n( )＝0，whenH n( )1. （7）

It is clear that ^{H n( )} is a completely multiplicative function as well. And

( ) ( )

H p H q 1. If n has prime factors other than ^{p q,} , then ^{H n( )0}. Clearly, | ( ) |Y n 

1

( )

n

i

H i





 , that is one can transform the number of valued “1” point of { ( )}H n to the partial sum of ^{H n( )}.

Theorem 5. Let p and q be primes. Let^{{ ( )}H n} be the sequence generated by transforming {A_n}according to the rule ER({ , })p q of Question 2.Then for kN, the following equality holds:

| (Y kpq) | | (Y kp) || (Y kq) || ( ) | 1Y k  （8）

Proof. Let nY kpq( ) satisfy H n( )1. Then n can be divided by p or q if n does not equal to 1.

Case 1: if p n| ，since H p( )1，it follows that ( )n ( )n ( ) ( ) 1

H H H p H n

p  p   .

Hence n ( ), ( )

Y kq n pY kq

p  ；

Case 2: If ^{q n|} ，since H q( )1，it follows that ( ) 1n

H q  ，n ( ), ( ) Y kp n qY kp

q  .

So, ^{Y kpq( ) pY kq( ) qY kp( ) {1}}, and ^{Y kpq( )pY kq( ) qY kp( ) {1}} is clear.

Hence ^{Y kpq( ) pY kq( ) qY kp( ) {1}}.

The above ^{pY kq( )} and ^{qY kp( )} stand for the sets ^{{pn n| Y kq( )}} and {qn n| Y kq( )} respectively. It is clear that

( ) ( ) { | ( )} ( )

pY kp qY kp  pqn n Y k  pqY k ，

| pY kq( ) | | ( Y kq) |，|qY kp( ) | | (Y kp) |，|pqY k( ) | | ( ) | Y k . By Principle of cross-classification, we get that

| (Y kpq) | | pY kq( ) ||qY kp( ) ||pY kq( ) qY kp( ) | 1

＝| (Y kq) || (Y kp) || ( ) | 1Y k  . □

Theorem 6. Let p and q be primes. Let^{{ ( )}H n} be the sequence generated by transforming {A_n}according to the rule ER^{({ , })p q} of Question 2.Then for kN, the following equality holds:

| (Y kp q^{r s}) | | ( Y kp^r) || (Y kq^s) | | ( ) | Y k rs （9）

Proof. Let number-theoretic function H n( ) be defined by（7），and we have | ( ) |Y n 

1

( )

n

i

H i





 ,（9）can be written as follows:

1

( )

r s

kp q

i

H i





 ^＝

1

( )

kpr

i

H i





 ⁺

1

( )

kqs

i

H i





 ^-

1

( )

k

i

H i rs





  . ^{( 9 )} We prove ^{(9 )} by induction in the following part. It follows from Theorem 5 that ^{(9 )} holds if r＝1，s＝1.

Assume that (9 ) holds for all r≤u, s≤v . If r＝u+1, s≤v, we have

1

1

( )

u s

kp q

i

H i







 ^{＝( )}

1

( )

kp pu

i

H i





 ^{+( )}

1

( )

kp qs

i

H i





 ^-

1

( )

kp

i

H i





 ^+us

＝

1

1

( )

kpu

i

H i







 ⁺

1 1 1

( ) ( ) ( ) 1

kp kqs k

i i i

H i H i H i s

  

 

      

 

 



  

^-

1

( )

kp

i

H i





 ^+us （unfold

( )

1

( )

kp qs

i

H i





 by induction hypothesis）

＝

1

1

( )

kpu

i

H i







 ⁺

1 1

( ) ( ) ( 1)

kqs k

i i

H i H i u s

 

     

 

^.

So, if r＝u+1, s≤v ,(9 ) holds. Similarly one can prove that if r≤u+1, s＝v+1, (9 ) holds as well. Hence the result follows from induction principle.

For the case k＝1，the following corollary holds：

Corollary3. Let p and q be primes. Let^{{ ( )}H n} be the sequence generated by transforming {A_n}according to the rule ER^{({ , })p q} of Question 2. Then for ^{r s, N}, the following equality holds:

| (Y p q^{r s}) | | (Y p^r) || (Y q^s) | rs 1

For the case p, q equal to 2, 3 respectively, the following interesting corollary holds：

Corollary4. Let p=2, q=3, { ( )}H n be the sequence generated by transforming {A_n} according to the rule ER^{({ , })p q} of question two. Then for ^{k r s, , N} , the following equalities hold:

(1)

2 3

1

( )

k r s

i

H i



 ^{＝k2(22r 1)(3}₄ ^{2s1)+ 2 3}

1 1 1

( ) ( ) ( )

r s

k k k

i i i

H i H i H i

  

 

  

（or

2 3

2

( )

r s

r

k

i k

H i





^{＝ 2(22 1)(32 1)}

4

r s

k  

+

3

( )

k s

i k

H i



 ^）

（or

2 3

3

( )

r s

s

k

i k

H i





^{＝ 2(22 1)(32 1)}

4

r s

k  

+

2

( )

k r

i k

H i



 ^）

(2)

2 3

2

( )

r s

i r

H i



 ^{＝(22 1)(32 1)}

4

r  s

+

3

1

( )

s

i

H i





（or

2 3

3

( )

r s

i s

H i



 ^{＝(22 1)(32 1)}

4

r  s 

+

2

1

( )

r

i

H i



 ^）

(3)

2 3

3 1

( )

s

i s

H i



 



^{＝32 1 1}

4

s 

,

3 2

2 1

( )

r

i r

H i



 



^＝22r1

(4)

6 3

2

( ) ( )

k k

i k i k

H i H i

 







^＝6k2,

18 9

2

( ) ( )

k k

i k i k

H i H i

 







^＝60k2

(5) | ( 2 3 ) | | ( 2 ) |Y k ^{r s} Y k ^r | ( 3 ) | | ( ) |Y k ^s  Y k rs

| (2 3 ) | | (2 ) |Y ^{r s}  Y ^r | (3 ) |Y ^s  rs 1

Section 3

In this section, we extend the problem to the situation including m arbitrary primes as follows.

Question 3: Given m primesp_i, i1, 2,...,m, we obtain a new sequence { ( )}H n by

transforming{A_n}according to the rule ER({ ,p p_{1 2},...,p_m}) as follows:

(1)For each i{1, 2,..., }m , if An has factor pi, then we substitute An with A_n/ p_i until it does not have factor pi

(2) If An does not have factor p i_i, 1, 2,...,m, then keep it unchanged.

What properties does { ( )}H n have?

The rule ER({ ,p p_{1 2},...,p_m}) above can also be written as follows:

If

1

,

i

m r

n i

i

A X p



 



^where

1

( , ) 1, 0, 1, 2,..., , 1

m

i i

i

X p r i m X



   



are integers, then

( ) H n X .

For instance, let the given primes be 2, 3,5, then the generated sequence ^{{ ( )}H n} is:

1,1,1,1,1,1,7,1,1,1,11,1,13,7,1,1,17,1,19,1,7,11,23,1,1,13,1,7,29,1,31,1,11,17,7,1,37,19, 13,1,41,7,43,11,1,23,47,1,49,…… （10）

The new sequence is more complicated. As in the previous section, we know H n( ) is a totally multiplicative function.

Let

1 m

i i

 p







^{,  l  / pl , l j, / p pl j,l j t, , / p p pl j t ,…;}

1

i

m r i i

 p







^,

/ ^rl

l pl

  ,_{l j,} / p p_l^{rl r}_j^j,_{l j t, ,}  / p p p_l^{rl r}_j^j _t^rt,…1 l m,1  l j m,… （11）

For the sum problem of { ( )}H n , we propose the following Theorem 7 and Theorem 8.

Theorem 7. Given m primes p p1, 2,...,p_m, let ^{{ ( )}H n} be the sequence generated by transforming{A_n}according to the rule ER({ ,p p_{1 2},...,p_m})of Question 3. ^Let   , _l, _{l j}, ,..._be described as (11), then for kN, the following equality holds:

1

( )

k

i

H i





 ^{＝ 2} 1

( 1)

2

m

i i

i

k p p





 ⁺

1 m



l

1

( )

k l

i

H i





 ^{- ,}

1 1

( )

k l j

l j m i

H i



  

 

 ^+…