For any positive integer n, let Sp(n) denotes the smallest positive integer such that Sp(n)! is divisible by p n, where p be a prime. The main purpose of this paper is using the elementary methods to study the relationship between Sp(n) and Sp(kn), and give an interesting identity

Weiyi Zhu

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Scientia Magna
Vol. 2 (2006), No. 4, 87-90
The relationship between Sp(n) and Sp(kn)
Weiyi Zhu
College of Mathematics, Physics and Information Engineer, Zhejiang Normal University
Jinhua, Zhejiang, P.R.China
Abstract For any positive integer n, let Sp(n) denotes the smallest positive integer such
that Sp(n)! is divisible by p
n, where p be a prime. The main purpose of this paper is using
the elementary methods to study the relationship between Sp(n) and Sp(kn), and give an
interesting identity.
Keywords The primitive numbers of power p, properties, identity
x1. Introduction and Results
Let p be a prime and n be any positive integer. Then we de¯ne the primitive numbers
of power p (p be a prime) Sp(n) as the smallest positive integer m such that m! is divided by
pn. For example, S3(1) = 3; S3(2) = 6; S3(3) = S3(4) = 9; ¢¢¢ : In problem 49 of book [1],
Professor F.Smarandache asked us to study the properties of the sequence fSp(n)g. About this
problem, Zhang Wenpeng and Liu Duansen [3] had studied the asymptotic properties of Sp(n),
and obtained an interesting asymptotic formula for it. That is, for any ¯xed prime p and any
positive integer n, they proved that
Sp(n) = (p ¡ 1)n + O
µ
p
lnp
lnn
¶
:
Yi Yuan [4] had studied the asymptotic property of Sp(n) in the form
1
p
X
n·x
jSp(n + 1) ¡ Sp(n)j,
and obtained the following result: for any real number x ¸ 2, let p be a prime and n be any
positive integer,
1
p
X
n·x
jSp(n + 1) ¡ Sp(n)j = x
µ
1 ¡
1
p
¶
+ O
µ
lnx
lnp
¶
:
Xu Zhefeng [5] had studied the relationship between the Riemann zeta-function and an
in¯nite series involving Sp(n), and obtained some interesting identities and asymptotic formulae
for Sp(n). That is, for any prime p and complex number s with Res > 1, we have the identity:
1 X
n=1
1
Ss
p(n)
=
³(s)
ps ¡ 1
;
where ³(s) is the Riemann zeta-function.88 Weiyi Zhu No. 4
And, let p be a ¯xed prime, then for any real number x ¸ 1 he got
1 X
n=1
Sp(n)·x
1
Sp(n)
=
1
p ¡ 1
µ
lnx + ° +
plnp
p ¡ 1
¶
+ O(x¡ 1
2+");
where ° is the Euler constant, " denotes any ¯xed positive number.
Chen Guohui [7] had studied the calculation problem of the special value of famous Smaran-
dache function S(n) = minfm : m 2 N; njm!g. That is, let p be a prime and k an integer with
1 · k < p. Then for polynomial f(x) = xnk +xnk¡1 +¢¢¢+xn1 with nk > nk¡1 > ¢¢¢ > n1, we
have:
S(pf(p)) = (p ¡ 1)f(p) + pf(1):
And, let p be a prime and k an integer with 1 · k < p, for any positive integer n, we have:
S
³
pkp
n´
= k
µ
Á(pn) +
1
k
¶
p;
where Á(n) is the Euler function. All these two conclusions above also hold for primitive function
Sp(n) of power p.
In this paper, we shall use the elementary methods to study the relationships between
Sp(n) and Sp(kn), and get some interesting identities. That is, we shall prove the following:
Theorem. Let p be a prime. Then for any positive integers n and k with 1 · n · p and
1 < k < p, we have the identities:
Sp(kn) = kSp(n), if 1 < kn < p;
Sp(kn) = kSp(n) ¡ p
·
kn
p
¸
, if p < kn < p2, where [x] denotes the integer part of x.
x2. Two simple Lemmas
To complete the proof of the theorem, we need two simple lemmas which stated as following:
Lemma 1. For any prime p and any positive integer 2 · l · p ¡ 1, we have:
(1) Sp(n) = np, if 1 · n · p;
(2) Sp(n) = (n ¡ l + 1)p, if (l ¡ 1)p + l ¡ 2 < n · lp + l ¡ 1.
Proof. First we prove the case (1) of Lemma 1. From the de¯nition of Sp(n) = minfm :
pnjm!g, we know that to prove the case (1) of Lemma 1, we only to prove that pnk(np)!. That is,
pnj(np)! and pn+1y(np)!. According to Theorem 1.7.2 of [6] we only to prove that
1 X
j=1
·
np
pj
¸
= n.
In fact, if 1 · n < p, note that
·
n
pi
¸
= 0; i = 1; 2;¢¢¢, we have
1 X
j=1
·
np
pj
¸
=
1 X
j=1
·
n
pj¡1
¸
= n +
·
n
p
¸
+
·
n
p2
¸
+ ¢¢¢ = n:
This means Sp(n) = np. If n = p, then
1 X
j=1
·
np
pj
¸
= n + 1, but pp y (p2 ¡ 1)! and ppjp2!. This
prove the case (1) of Lemma 1. Now we prove the case (2) of Lemma 1. Using the same methodVol. 2 The relationship between Sp(n) and Sp(kn) 89
of proving the case (1) of Lemma 1 we can deduce that if (l ¡1)p+l ¡2 < n · lp+l ¡1, then
·
n ¡ l + 1
p
¸
= l ¡ 1;
·
n ¡ l + 1
pi
¸
= 0; i = 2;3;¢¢¢ :
So we have
1 X
j=1
·
(n ¡ l + 1)p
pj
¸
=
1 X
j=1
·
n ¡ l + 1
pj¡1
¸
= n ¡ l + 1 +
·
n ¡ l + 1
p
¸
+
·
n ¡ l + 1
p2
¸
+ ¢¢¢
= n ¡ l + 1 + l ¡ 1 = n:
From Theorem 1.7.2 of reference [6] we know that if (l ¡ 1)p + l ¡ 2 < n · lp + l ¡ 1, then
pnk((n ¡ l + 1)p)!. That is, Sp(n) = (n ¡ l + 1)p. This proves Lemma 1.
Lemma 2. For any prime p, we have the identity Sp(n) = (n¡p+1)p, if p2¡2 < n · p2.
Proof. It is similar to Lemma 1, we only need to prove pnk((n ¡ p + 1)p)!. Note that if
p2 ¡ 2 < n · p2, then
·
n ¡ p + 1
p
¸
= p ¡ 1;
·
n ¡ p + 1
pi
¸
= 0; i = 2; 3;¢¢¢. So we have
1 X
j=1
·
(n ¡ p + 1)p
pj
¸
=
1 X
j=1
·
n ¡ p + 1
pj¡1
¸
= n ¡ p + 1 +
·
n ¡ p + 1
p
¸
+
·
n ¡ p + 1
p2
¸
+ ¢¢¢
= n ¡ p + 1 + p ¡ 1 = n:
From Theorem 1.7.2 of [6] we know that if p2 ¡ 2 < n · p2, then pnk((n ¡ p + 1)p)!. That is,
Sp(n) = (n ¡ p + 1)p. This completes the proof of Lemma 2.
x3. Proof of Theorem
In this section, we shall use above Lemmas to complete the proof of our theorem.
Since 1 · n · p and 1 < k < p, therefore we deduce 1 < kn < p2. We can divide 1 < kn <
p2 into three interval 1 < kn < p, (m ¡ 1)p + m ¡ 2 < kn · mp + m ¡ 1 (m = 2; 3;¢¢¢ ;p ¡ 1)
and p2 ¡ 2 < kn · p2. Here, we discuss above three interval of kn respectively:
i) If 1 < kn < p, from the case (1) of Lemma 1 we have
Sp(kn) = knp = kSp(n):
ii) If (m ¡ 1)p + m ¡ 2 < kn · mp + m ¡ 1 (m = 2;3;¢¢¢ ;p ¡ 1), then from the case (2)
of Lemma 1 we have
Sp(kn) = (kn ¡ m + 1)p = knp ¡ (m ¡ 1)p = kSp(n) ¡ (m ¡ 1)p:
In fact, note that if (m ¡ 1)p + m ¡ 2 < kn < mp + m ¡ 1 (m = 2;3;¢¢¢ ;p ¡ 1), then
m ¡ 1 +
·
m ¡ 2
p
¸
<
·
kn
p
¸
< m +
·
m ¡ 1
p
¸
. Hence,
·
kn
p
¸
= m ¡ 1. If kn = mp + m ¡ 1,90 Weiyi Zhu No. 4
then
·
kn
p
¸
= m, but pmp+m¡1 y ((mp + m ¡ 1)p ¡ 1)! and pmp+m¡1j((mp + m ¡ 1)p)!. So we
immediately get
Sp(kn) = kSp(n) ¡ p
·
kn
p
¸
:
iii) If p2 ¡ 2 < kn · p2, from Lemma 2 we have
Sp(kn) = (kn ¡ p + 1)p = knp ¡ (p ¡ 1)p:
Similarly, note that if p2 ¡2 < kn · p2, then p¡
·
2
p
¸
<
·
kn
p
¸
· p. That is,
·
kn
p
¸
= p¡1. So
we may immediately get
Sp(kn) = kSp(n) ¡ p
·
kn
p
¸
:
This completes the proof of our Theorem.
References
[1] F. Smarandache, Only problems, not solutions, Chicago, Xiquan Publishing House,
1993.
[2] Tom M. Apostol, Introduction to analytic number theory, New York, Springer-Verlag,
1976.
[3] Zhang Wenpeng and Liu Duansen, On the primitive numbers of power p and its asymp-
totic property, Smarandache Notions Journal, 13(2002), pp. 173-175.
[4] Yi Yuan, On the primitive numbers of power p and its asymptotic property, Scientia
Magna, 1(2005), No. 1, pp. 175-177.
[5] Xu Zhefeng, Some arithmetical properties of primitive numbers of power p, Scientia
Magna, 2(2006), No. 1, pp. 9-12.
[6] Pan Chengdong and Pan Chengbiao, The elementary number theory, Beijing University
Press, Beijing, 2003.
[7] Chen Guohui, Some exact calculating formulas for the Smarandache function, Scientia
Magna, 2(2006), No. 3, pp. 95-97.

The relationship between Sp(n) and Sp(kn)

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