By using the afore mentioned bounds, we study the equationvp(n

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Volume 6, Issue 2, Article 29, 2005

EQUATIONS AND INEQUALITIES INVOLVING v_p(n!)

MEHDI HASSANI DEPARTMENT OFMATHEMATICS

INSTITUTE FORADVANCEDSTUDIES INBASICSCIENCES

ZANJAN, IRAN

mhassani@iasbs.ac.ir

Received 14 June, 2004; accepted 16 February, 2005 Communicated by L. Tóth

Dedicated to Professor Bahman Mehri on the occasion of his 70th birthday

ABSTRACT. In this paper we studyvp(n!), the greatest power of primepin factorization ofn!.

We find some lower and upper bounds forvp(n!), and we show thatvp(n!) = _p−1ⁿ +O(lnn).

By using the afore mentioned bounds, we study the equationv_p(n!) = v for a fixed positive integer v. Also, we study the triangle inequality aboutv_p(n!), and show that the inequality p^vp(n!)> q^vq(n!)holds for primesp < qand sufficiently large values ofn.

Key words and phrases: Factorial function, Prime number, Inequality.

2000 Mathematics Subject Classification. 05A10, 11A41, 26D15, 26D20.

1. INTRODUCTION

As we know, for everyn ∈N,n! = 1×2×3× · · · ×n. Letv_p(n!)be the highest power of primepin factorization ofn!to prime numbers. It is well-known that (see [3] or [5])

(1.1) v_p(n!) =

∞

X

k=1

n p^k

= [^lnlnⁿp]

X

k=1

n p^k

,

in which [x]is the largest integer less than or equal to x. An elementary problem aboutn!is finding the number of zeros at the end of it, in which clearly its answer isv₅(n!). The inverse of this problem is very nice; for example finding values ofnin whichn!terminates in 37 zeros [3], and generally finding values ofn such thatv_p(n!) = v. We show that if v_p(n!) = v has a

ISSN (electronic): 1443-5756

c 2005 Victoria University. All rights reserved.

I deem my duty to thank A. Abedin-Zade and Y. Roodghar-Amoli for their comments on the Note and Problem 1.

119-04

solution then it has exactlypsolutions. For doing these, we need some properties of[x], such as

(1.2) [x] + [y]≤[x+y] (x, y ∈R),

and (1.3)

hx n i

= [x]

n

(x∈R, n∈N).

2. ESTIMATINGv_p(n!)

Theorem 2.1. For everyn∈Nand primep, such thatp≤n, we have:

(2.1) n−p

p−1 − lnn

lnp < v_p(n!)≤ n−1 p−1. Proof. According to the relation (1.1), we havev_p(n!) =Pm

k=1

hn p^k

i

in whichm =h

lnn lnp

i , and sincex−1<[x]≤x, we obtain

n

m

X

k=1

1

p^k −m < v_p(n!)≤n

m

X

k=1

1 p^k, consideringPm

k=1 1 p^k = ¹⁻

1 pm

p−1 , we obtain n

p−1

1− 1 p^m

−m < v_p(n!)≤ n p−1

1− 1

p^m

,

and combining this inequality with ^ln_lnⁿ_p −1< m≤ ^ln_lnpⁿ completes the proof.

Corollary 2.2. For everyn∈Nand primep, such thatp≤n, we have:

v_p(n!) = n

p−1 +O(lnn).

Proof. By using (2.1), we have

0<

n

p−1 −v_p(n!) lnn < 1

lnp+O 1

lnn

,

and this yields the result.

Note that the above corollary asserts thatn!ends approximately in ⁿ₄ zeros [1].

Corollary 2.3. For everyn ∈Nand primep, such thatp≤n, and for alla∈(0,∞)we have:

(2.2) n−p

p−1 − 1 lnp

n

a + lna−1

< v_p(n!).

Proof. Consider the functionf(x) = lnx. Since,f⁰⁰(x) =−_x¹2,lnxis a concave function and so, for everya∈(0,+∞)we have

lnx≤lna+ 1

a(x−a),

combining this with the left hand side of (2.1) completes the proof.

3. STUDY OF THE EQUATIONv_p(n!) =v

Supposev ∈Nis given. We are interested in finding the values ofnsuch that in factorization ofn!, the highest power ofp, is equal tov. First, we find some lower and upper bounds for these n’s.

Lemma 3.1. Supposev ∈Nandpis a prime andvp(n!) =v, then we have

(3.1) 1 + (p−1)v ≤n < v+_p−1^p +ln(1+(p−1)v) lnp −_ln¹_p

1

p−1 − (1+(p−1)v) ln¹ p

.

Proof. For proving the left hand side of (3.1), use right hand side of (2.1) with the assumption v_p(n!) =v, and for proving the right hand side of (3.1), use (2.2) witha= 1 + (p−1)v.

Lemma 3.1 suggests an interval for the solution ofv_p(n!) = v. In the next lemma we show that it is sufficient for one to check only multiples ofpin above interval.

Lemma 3.2. Supposem∈Nandpis a prime, then we have

(3.2) v_p((pm+p)!)−v_p((pm)!)≥1.

Proof. By using (1.1) and (1.2) we have

v_p((pm+p)!) =

∞

X

k=1

pm+p p^k

≥

∞

X

k=1

pm p^k

+

∞

X

k=1

p p^k

= 1 +v_p((pm)!),

and this completes the proof.

In the next lemma, we show that ifv_p(n!) =v has a solution, then it has exactlypsolutions.

In fact, the next lemma asserts that if v_p((mp)!) = v holds, then for all 0 ≤ r ≤ p− 1, v_p((mp+r)!) = valso holds.

Lemma 3.3. Supposem∈Nandpis a prime, then we have

(3.3) v_p((m+ 1)!)≥v_p(m!),

and

(3.4) v_p((pm+p−1)!) =v_p((pm)!).

Proof. For proving (3.3), use (1.1) and (1.2) as follows

v_p((m+ 1)!) =

∞

X

k=1

m+ 1 p^k

≥

∞

X

k=1

m p^k

+

∞

X

k=1

1 p^k

=

∞

X

k=1

m p^k

=v_p(m!).

For proving (3.4), it is enough to show that for allk ∈N,h_pm+p−1

p^k

i

=h

pm p^k

i

and we do this by induction onk; fork = 1, clearlyh

pm+p−1 p

i

=h

pm p

i

. Now, by using (1.3) we have

pm+p−1 p^k+1

=

"_pm+p−1

p^k

p

#

=





hpm+p−1 p^k

i

p



=



 hpm

p^k

i

p





=

"_pm

p^k

p

#

= pm

p^k+1

.

This completes the proof.

So, we have proved that

Theorem 3.4. Suppose v ∈ N and p is a prime. For solving the equation v_p(n!) = v, it is sufficient to check the valuesn=mp, in whichm ∈Nand

(3.5)

1 + (p−1)v p

≤m≤

"

v+ _p−1^p + ln(1+(p−1)v) lnp − _ln¹_p

p

p−1 −(1+(p−1)v) lnp^p

# .

Also, ifn=mpis a solution ofv_p(n!) =v, then it has exactlypsolutionsn=mp+r, in which 0≤r≤p−1.

Note and Problem 1. As we see, there is no guarantee of the existence of a solution forvp(n!) = v. In fact we need to show that {vp(n!)|n ∈ N} = N; however, computational observations suggest thatn=p

1+(p−1)v p

usually is a solution, such that||x||is the nearest integer tox, but we cannot prove it.

Note and Problem 2. Other problems can lead us to other equations involving v_p(n!); for example, supposen, v∈Ngiven, find the value of primepsuch thatv_p(n!) =v.

Or, supposepandqare primes andf :N² →Nis a prime value function, for whichn’s do we havevp(n!) +vq(n!) =vf(p,q)(n!)? And many other problems!

4. TRIANGLE INEQUALITYCONCERNINGv_p(n!) In this section we are going to comparev_p((m+n)!)andv_p(m!) +v_p(n!).

Theorem 4.1. For everym, n∈Nand primep, such thatp≤min{m, n}, we have

(4.1) v_p((m+n)!)≥v_p(m!) +v_p(n!), and

(4.2) v_p((m+n)!)−v_p(m!)−v_p(n!) =O(ln(mn)).

Proof. By using (1.1) and (1.2), we have

v_p((m+n)!) =

∞

X

k=1

m+n p^k

≥

∞

X

k=1

m p^k

+

∞

X

k=1

n p^k

=v_p(m!) +v_p(n!).

Also, by using (2.1) and (4.1) we obtain

0≤v_p((m+n)!)−v_p(m!)−v_p(n!)

< 2p−1

p−1 + ln(mn)

lnp ≤3 + ln(mn) ln 2 ,

this completes the proof.

More generally, ifn₁, n₂, . . . , n_t ∈Nandpis a prime, in whichp≤min{n₁, n₂, . . . , n_t}, by using an extension of (1.2), we obtain

v_{p t}

X

k=1

n_k

!

!

!

≥

t

X

k=1

v_p(n_k!),

and by using this inequality and (2.1), we obtain

0≤v_{p t}

X

k=1

n_k

!

!

!

−

t

X

k=1

v_p(n_k!)

< tp−1

p−1 + ln(n₁n₂· · ·n_t) lnp

≤2t−1 + ln(n₁n₂· · ·n_t) ln 2 , and consequently we have

v_{p t}

X

k=1

n_k

!

!

!

−

t

X

k=1

v_p(n_k!) =O(ln(n₁n₂· · ·n_t)).

Note and Problem 3. Suppose f : N^t → N is a function and p is a prime. For which n₁, n₂, . . . , n_t∈N, do we have

v_p((f(n₁, n₂, . . . , n_t)!)≥f(v_p(n₁!), v_p(n₂!), . . . , v_p(n_t!))?

Also, we can consider the above question in other view points.

5. THEINEQUALITYp^vp(n!) > q^vq(n!)

Suppose p and q are primes and p < q. Since v_p(n!) ≥ v_q(n!), comparing p^vp(n!) and q^vq(n!) becomes a nice problem. In [2], by using elementary properties about[x], the inequal- ity p^vp(n!) > q^vq(n!) was considered for some special cases. In addition, it was shown that 2^v2(n!) >3^v3(n!)holds for alln≥4. In this section we studyp^vp(n!) > q^vq(n!) in the more general case and also reprove2^v2(n!) >3^v3(n!).

Lemma 5.1. Supposepandqare primes andp < q, then

p^q−1 > q^p−1.

Proof. Consider the function

f(x) = x^x−11 (x≥2).

A simple calculation yields that forx≥2we have

f⁰(x) =−x^2−xx−1(xlnx−x+ 1) (x−1)² <0,

so,f is strictly decreasing andf(p)> f(q). This completes the proof.

Theorem 5.2. Supposepandqare primes andp < q, then for sufficiently largen’s we have

(5.1) p^vp(n!) > q^vq(n!).

Proof. Sincep < q, Lemma 5.1 yields that ^p_q^q−1p−1 > 1 and so, there exitsN ∈ N such that for n > N we have

p^q−1 q^p−1

n

≥ p^p(q−1)

q^p−1 n^{(p−1)(q−1)}. Thus,

p^n(q−1)

n^{(p−1)(q−1)}p^p(q−1) ≥ q^n(p−1) q^p−1 , and therefore,

p^p−1n

np^p−1p ≥ q^q−1n q^q−11 . So, we obtain

p^{n−pp−1−lnlnnp} ≥q^n−1q−1,

and considering this inequality with (2.1), completes the proof.

Corollary 5.3. Forn= 2andn ≥4we have

(5.2) 2^v2(n!) >3^v3(n!).

Proof. It is easy to see that forn≥30we have 4

3 n

≥ 16 3 n²,

and by Theorem 5.2, we yield (5.2) forn ≥ 30. For n = 2and4 ≤ n < 30check it using a

computer.

A Computational Note. In Theorem 5.2, the relation (5.1) holds forn > N (see its proof). We can check (5.1) forn≤N at most by checking the following number of cases:

R(N) := #{(p, q, n)|p, q ∈P, n= 3,4, . . . , N, andp < q ≤N}, in whichPis the set of all primes. If,π(x) =The number of primes≤x, then we have

R(N) =

N

X

n=3

#{(p, q)|p, q ∈P, andp < q ≤n}= 1 2

N

X

n=3

π(n)(π(n)−1).

But, clearlyπ(n)< nand this yields that

R(N)< N³ 6 .

Of course, we have other bounds forπ(n)sharper thannsuch as [4]

π(n)≤ n lnn

1 + 1

lnn + 2.25 ln²n

(n ≥355991),

and by using this bound we can find sharper bounds forR(N).

REFERENCES

[1] A. ADLERANDJ.E. COURY, The Theory of Numbers, Bartlett Publishers, 1995.

[2] I. B ˘AL ˘ACENOIU, Remarkable inequalities, Proceedings of the First International Conference on Smarandache Type Notions in Number Theory (Craiova, 1997), 131–135, Am. Res. Press, Lupton, AZ, 1997.

[3] D.M. BURTON, Elementary Number Theory (Second Edition), Universal Book Stall, 1990.

[4] P. DUSART, Inégalités explicites pour ψ(X), θ(X), π(X) et les nombres premiers, C. R. Math.

Acad. Sci. Soc. R. Can., 21(2) (1999), 53–59.

[5] M.B. NATHANSON, Elementary Methods in Number Theory, Springer, 2000.