# Further Results on a Curious Arithmetic Function

Authors: Long Chen; Kaimin Cheng; Tingting Wang
Journal: Journal of Mathematics
 (2020)

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## Abstract

Letp be an odd prime number and n be a positive integer. Let vpn, N∗, and Q+ denote the p-adic valuation of the integer n, the set of positive integers, and the set of positive rational numbers, respectively. In this paper, we introduce an arithmetic function fp:N∗⟶Q+ defined by fpn≔n/pvpn1−vpn for any positive integer n. We show several interesting arithmetic properties about that function and then use them to establish some curious results involving the p-adic valuation. Some of these results extend Farhi’s results from the case of even prime to that of odd prime.


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## Body

## 1. Introduction

Letp be any given prime and n be a positive integer. As usual, let vpn, N∗=N∖0, and Q+=x∈Qx>0 denote the p-adic valuation of the integer n, the set of positive integers, and the set of positive rational numbers, respectively. Arithmetic functions are classical topics in number theory (see, for example, [1–3]). Recent years, some kinds of periodic functions were introduced by some authors for the investigation of the least common multiple of integer sequences (see, for instance, [4–9]). In this paper, we introduce an interesting arithmetic function fp:N∗⟶Q+ defined by(1)fpn≔npvpn1−vpn,for any positive integer n. We have, for example, f51=1,f55=1,f550=1/2,f575=1/3,f5250=1/4 etc. It is clear that fpn is not always an integer for any given prime p.Throughout this paper, letp always denote an odd prime. In this paper, we present several interesting arithmetic properties about that function and then use them to establish some curious results involving the p-adic valuation. This paper is organised as follows. First in Section 2, we show that the product Fpn≔∐r=1nfpr is always an integer for any positive integer n, and it is, in fact, a multiple of all prime number not equal to p and not exceeding n. Subsequently in Section 3, we first give a lower bound for Hpn defined in Section 2. Also, two properties of p-adic valuation are provided. From these results, we can show that limn⟶∞Fpn1/n does not exist for the odd prime p, which is very different from the case of p=2 since Farhi [10] proved that limn⟶∞F2n1/n=4. Finally, Section 4 is devoted to derive an upper bound of Fpn if p≥3. One notes that f2n was studied by Farhi [10], and f3n was investigated by Wang et al. [11]. Actually, many results they obtained would be extended in this paper to the case of all odd primes p.In order to investigate the functionfp, two auxiliary arithmetic functions g:Q+⟶N∗ and hp:N∗⟶Q+ are needed, defined, respectively, by(2)gx≔x,ifx∈N∗,1,otherwise,(3)hpr≔r∏i=1∞gr/pi,for any integer r∈N∗. Note that the function hp is well defined since the product in the denominator of the right-hand side of (3) is actually finite because gr/pi=1 for any sufficiently large i.Letx be a rational number. By vpx, denote the p-adic valuation of x. Define Npx by(4)Npx≔xpvpx.So,fpn=Npn1−vpn for any positive integer n. As usual, let ⋅ be the integer-part function, i.e., x represents the largest integer no more than x, and ⋅⋅ represent the binomial coefficient. Here, we give a well-known fact, which will be used frequently in this paper, as follows.Fact 1.
Each of following is true.(i)
x/a/b=x/ab for any a,b∈N∗ and x∈Q.(ii)
(a)
emm!<m+1m+1.(b)
m+1kk<m+1m+1/mmk.(c)
mk<mk/k! hold for any positive integers m and k.(iii)
(Legendre’s formula) Letp be a prime and n be a positive integer. Then,(5)vpn!=∑i=1∞npi=n−∑i=0saip−1,where ∑i=0saipi is the p-adic representation of n.


## 2. The Integrity of the Product Function

LetHpn≔∏r=1nhpr and Ln≔lcm1,…,n for any positive integer n. Now, we give the first result as follows.Theorem 1.
Letp be any given prime number and n be a positive integer. Then, the value Fpn is an integer.Proof.
First, by the definition ofg, for any positive integer r, we have(6)∏i=1∞grpi=∏i=1∞gpvpr−iNpr=∏i=1vprgpvpr−iNpr=∏i=1vprpvpr−iNpr=∏i=0vpr−1piNpr=pvprvpr−1/2Nprvpr.
Then,(7)hpr=r∏i=1∞gr/pi=pvpr3−vpr/2Npr1−vpr=pvpr3−vpr/2fpr.
It follows that(8)fpr=pvprvpr−3/2hpr.
So, for all positive integersn, we have(9)Fpn=p∑r=1nvprvpr−3/2Hpn.
Now, by applying the functionNp into each of two sides of (9), we obtain(10)Fpn=NpFpn=NpHpn.
Therefore, to prove Theorem1, we only need to prove that Hpn is an integer for any positive integer n. For this purpose, by the definition of g, one notes that ∏i=1rgi/a=r/a! for any positive integers r and a. Hence,(11)Hpn=∏r=1nr∏i=1∞gr/pi=n!∏i=1∞∏r=1ngr/pi=n!∏i=1∞n/pi!.
Additionally,∑i=1∞n/pi≤∑i=1∞n/pi=n/p−1≤n. Thus, n!/∏i=1∞n/pi! is a multiple of the multinomial coefficient n/p+n/p2+⋯n/p,n/p2,… which is an integer. So, n!/∏i=1∞n/pi! must be an integer. This finishes the proof of Theorem 1.
Next, let us discuss the divisibility ofFpn by NpLn. In fact, we have the following result.Theorem 2.
For any positive integern, Fpn is a multiple of NpLn. In particular, Fpn is a multiple of all prime numbers not equal to p and not exceeding n.Proof.
By (10) and (11), we know that to prove Theorem 2, it is sufficient to show that n!/∏i=1∞n/pi! is a multiple of Ln, which is equivalent to prove that for all prime numbers q not exceeding n, we have(12)vqn!∏i=1∞n/pi!≥αq,where αq=vqLn. On the one hand, by (i) and (iii) in Fact 1, one has(13)vqn!∏i=1∞n/pi!=∑i=1∞nqi−∑i=1∞∑j=1∞n/qipj=∑i=1∞n/qi−∑j=1∞n/qipj=∑i=1αqnqi−∑j=1∞n/qipj.
On the other hand, asq≤n, for each 1≤i≤αq, one then gets(14)nqi−∑j=1∞n/qipj≥nqi−∑j=1∞n/qipj=nqi1−∑j=1∞1pj=p−2p−1nqi>0.
Besides,n/qi−∑j=1∞n/qi/pj is an integer. It then infers that(15)nqi−∑j=1∞n/qipj≥1.
Therefore by (13) and (15), we have(16)vqn!∏i=1∞n/pi!≥∑i=1αq1=αq,which implies that (12) is true. So, the proof of Theorem 2 is complete.


## 3. Properties ofHpn and Fpn

Letp be any odd prime. Now, we turn our attention to the lower bound for Hpn. In this section, we first obtain a lower bound for Hpn. Secondly, two properties of p-adic valuations will be given. Finally, one then concludes that limn⟶∞Fpn1/n does not exist, which varies from the result of the case p=2 obtained by Farhi [10] that limn⟶∞F2n1/n=4. First of all, we present one lemma as follows.Lemma 1.
Letp be a prime number. Let mandr be integers with m≥1 and 1≤r≤p−1. Then, for any positive integer i, one has(17)mp+rpi=mppi.Proof.
We prove (17) by induction on i as follows.Fori=1, (17) is clearly true.For the given integeri≥2, suppose that (17) holds for any integer j<i. Let us now show that (17) is true for the integer i. On the one hand, by Fact 1, we have(18)mp+rpi=mp+r/pi−1p.
On the other hand, by the assumption, we have(19)mp+rpi−1=mppi−1.
So, combining (18) with (19), it then follows from Fact 1 that(20)mp+rpi=mppi,as desired. This completes the proof of Lemma 1.
Letp≥3 be a given prime number and n be any positive integer. Now, we present a lower bound for Hpn.Theorem 3.
Letp be an odd prime number and n be a positive integer. Then, we have(21)Hpn≥logpnn.Proof.
Letn be positive integer. First, we claim that(22)Hp1n≤Hp2n,for all primes p1,p2 with p1<p2. In fact, for any positive integer i, one has(23)np1i≥np2i.
It follows that(24)np1i!≥np2i!,which infers that(25)∏i=1∞np1i!≥∏i=1∞np2i!.
By (11), the desired result (22) follows. Now, let us continue to prove inequality (21). It is well known that log3n≥logpn for each positive integer n and each prime p≥3. Hence, for proving inequality (21), by (22), we only need to prove(26)H3n=n!∏i=1∞n/3i!≥log3nn,which will be done in what follows.
By using a computer, we have checked that (26) is true for 1≤n≤310=59049. Now, we show that (26) holds for any integer n>310 by induction on n. Let n>310 be the fixed positive integer. Assume that (26) is true for all integers less than n. We need to show that (26) is true for the integer n. We divide our proof into the following three cases.Case 1.
n=3m, where m is an integer with m>39. In this case, we have(27)H3n=H33m=3m!∏i=1∞3m/3i!=3m!m!⋅m!⋅m!∏i=1∞m/3i!=3m!m!⋅m!H3m.
Then, by the assumption thatH3m≥log3mm≥1 and the facts that 3m!/m!⋅m!≥2mm2mm≥2mm, we have(28)H3n≥2mm=2×3log3mm>log3m+13m=log33m3m=log3nn,since 2×3t>t+13 holds for any real number t>9. So, (26) is true if 3n.Case 2.
n=3m+1, where m is an integer with m≥39. Then, by Lemma 1 and the assumption, we derive that(29)H3n=H33m+1=3m+1!∏i=1∞3m+1/3i!=3m+13m!∏i=1∞3m/3i!=3m+1H33m>3m+1log33m3m.
But one claims that(30)inf3≤t<+∞t+1log3ttlog3t+1t+1>1.
In fact, letht≔t+1log3tt/log3t+1t+1 and h′t be the derivative function of ht. Then, one checks that h′t>0 for all real numbers t≥3. It follows that ht increases monotonically over the interval 3,+∞. Then, ht≥h3=1.577669…. So, (30) holds. Hence, from (29) and (30), we have that(31)H3n>log33m+13m+1=log3nn.
This means that (26) is true if 3n−1.Case 3.
n=3m+2, where m is an integer with m≥39. In this case, similar to Case 2, we have(32)H3n=3m+2H33m+1>log33m+23m+2=log3nn.
This tells us that (26) holds if 3n−2.
Combining the above cases, we have that (26) is true for the integer n. So, by mathematical induction, we have that (26) holds for any integer n>310. This completes the proof of Theorem 3.
In the following, we present two curious properties ofp-adic valuations from fpn.Theorem 4.
Letpandq be distinct primes and n be a positive integer. Then, we have(33)∑i=1nvpivqi≤∑i=1nvqi−logqn.Proof.
On the one hand, by Theorem2, we have(34)∑i=1nvqfpi=vqFpn≥vqNpLn=vqLn=logqn.
On the other hand, for any integeri≥1, we get(35)vqfpi=vqNpi1−vpi=1−vpivqNpi=1−vpivqi.
Therefore,(36)∑i=1n1−vpivqi≥logqn.
So, the desired result follows. This finishes the proof of Theorem4.Theorem 5.
Letp be a prime number and n be a positive integer. Let a0+a1p+a2p2+⋯+asps be the p-adic expansion of n. Then, we have(37)∑i=1nvpi3−vpi=2p−4p−12∑i=0spi−1ai+2p−1∑i=0siai.Proof.
First, it is easy to see that (37) is true when n<p. So, in what follows, we let n≥p. Then, (37) is equivalent to(38)∑i=1nvpi3−vpi=2p−4p−12∑i=1spi−1ai+2p−1∑i=1siai.
Now, we show that (38) holds. Taking the p-adic valuation on both sides of identity (9), using Fact 1 and (11), one has(39)∑i=1nvpi3−vpi2=vpHpn=vpn!∏i=1∞n/pi!=∑i=1∞npi−∑i=1∞∑j=1∞npi+j.
One notes that#i,j∈N∗2i+j=k=k−1, where k is an integer with k≥2. Then,(40)∑i=1∞∑j=1∞npi+j=∑k=2∞∑i+j=k∞npk=∑k=2∞k−1npk.
Then, (39) becomes(41)∑i=1nvpi3−vpi2=∑i=1∞npi−∑i=1∞inpi+1.
Since(42)p−1∑i=1∞i−1npi−inpi+1=0,it then follows that(43)∑i=1∞npi−∑i=1∞inpi+1=∑i=1∞p−1i−p+2npi−pinpi+1.
However, by thep-adic expansion of n: n=∑i=jsajpj, one has(44)p−1i−p+2npi−pinpi+1=p−2i−1npi+iai,if1≤i≤s,0,ifi>s.
Then, (43) together with (44) gives us that(45)∑i=1∞npi−∑i=1∞inpi+1=∑i=1∞p−2inpi+1+∑i=1siai,which infers that(46)∑i=1∞inpi+1=1p−1∑i=1∞npi−1p−1∑i=1siai.
It then follows from (iii) ofFact 1 that(47)∑i=1∞npi−∑i=1∞inpi+1=p−2p−1∑i=1∞npi+1p−1∑i=1siai=p−2p−12n−∑i=1sai+1p−1∑i=1siai=p−2p−12∑i=1spi−1ai+1p−1∑i=1siai.
Therefore, the desired result follows immediately from (41) and (47). The proof of Theorem 5 is complete.
Now, at the end of this section, by using Theorems3 and 5, one gets a result of limitation concerning Fpn as follows.Corollary 1.
Letp be an odd prime number. We have that(48)limn⟶∞Fpn1/n,does not exist.Proof.
Letp be an odd prime and n be any positive integer. First, by (9) and Theorems 3 and 5, one has(49)Fpn=p∑r=1nvprvpr−3/2Hpn=Hpnpp−2/p−12∑i=0spi−1ai+1/p−1∑i=0siai≥logpnnpp−2/p−12∑i=0spi−1ai+1/p−1∑i=0siai,where a0+a1p+a2p2+⋯+asps is the p-adic representation of n. Now, taking n=pm with m being a positive integer, one derives that(50)Fpn1/n=Fppm1/pm≥logppmpmpp−2/p−12pm−p−2/p−12+m/p−11/pm=mpp−2/p−12p−p−2/p−121/pmp1/p−1m/pm⟶+∞m⟶+∞.
It then follows that(51)limm⟶+∞Fppm1/pm=+∞.
So,limn⟶+∞Fpn1/n does not exist.


## 4. An Upper Bound forFpn

Letp be an odd prime. In this section, we give an upper bound for Fpn as follows.Theorem 6.
Letp be an odd prime number and n be a positive integer. We have that(52)Fpn<pnp−2n4n.Proof.
First, we show that(53)Hpn<pnp−2n4n,for all positive integers and odd primes p. This will be done by dividing into two cases.Case 4.
n<p. By the definition of Hpn and (ii) of Fact 1, one has(54)Hpn=∏r=1nhpr=n!<n+1n+1en.
So, in order to prove (53) holds for any prime p≥3, we only need to confirm that(55)n+1n+1en≤pnp−2n4n,for p≥3 in the case of n<p, which will be easily done by taking logarithm into both sides of above inequality.Case 5.
n≥p. Let us prove (53) by induction on n. First, it is easy to check (53) holds for n=p. Assume (53) is true for the integer less than n. Write n=pm+i, where m and i are integers with m≥1 and 0≤i≤p−1. It then follows from (11) and Lemma 1 that(56)Hpn=n!∏r=1∞n/pr!=pm+i!∏r=1∞pm+i/pr!=pm+i!∏r=1∞pm/pr!=p−2m+i!×pm+i!m!m!p−2m+i!m!∏r=1∞m/pr!=p−2m+i!pm+im,m,p−2m+iHpm=p−2m+i!2mmpm+ip−2m+iHpm.
Hence, by utilizing inequalities in (ii) ofFact 1, one derives that(57)Hpn<p−2m+i!4mpm+ip−2m+ip−2m+i!Hpm=4mpm+ip−2m+iHpm.
So, (57) gives us from the assumption that(58)Hpn<4mpm+ip−2m+ipmp−2m4m<42mn2p−2m+i<4nnpp−2n.
It follows that (53) holds for all integers n with n≥p. Therefore, (53) is proved. Now, by using (10) and (53), one arrives at(59)Fpn≤Hpn<pnp−2n4n,as desired. So, it completes the proof of Theorem 6.



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*Source: 1894162-2020-08-25.xml*