# The product of divisors minimum and maximum functions.

1. Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] denote the product of all divisors of n. The product-of- divisors minimum, resp. maximum functions will be defined by

T(n) = min{k [greater than or equal to] 1 : n|T(k)} (1)

and

T*(n) = max{k [greater than or equal to] 1 : T(k)|n}. (2)

There are particular cases of the functions [F.sup.A.sub.f], [G.sup.A.sub.g] defined by

F.sup.A.sub.f](n) = min{k [member of] A : n|f(k)}, (3)

and its "dual"

[G.sup.A.sub.g](n) = max{k [member of] A : g(k)|n}, (4)

where A [subset] N* is a given set, and f, g : N* [right arrow N are given functions, introduced in [8] and [9]. For A = N*, f (k) = g(k) = k! one obtains the Smarandache function S(n), and its dual S*(n), given by

S(n) = min{k [greater than or equal to] 1 : n|k!} (5)

and

S*(n) = max{k [greater than or equal to] 1 : k!|n}. (6)

The function S*(n) has been studied in [8], [9], [4], [1], [3]. For A = N*, f (k) = g(k) = [phi](k), one obtains the Euler minimum, resp. maximum functions

E(n) = min{k [greater than or equal to] 1 : n|[phi](k)} (7)

studied in [6], [8], [13], resp., its dual

E*(n) = max{k [greater than or equal to] 1 : [phi](k)|n}, (8)

studied in [13].

For A = N*, f (k) = g(k) = S(k) one has the Smarandache minimum and maximum functions

[S.sub.min](n) = min{k [greater than or equal to] 1 : n|[sigma](k)}, (9)

\[S.sub.max](n) = max{k [greater than or equal to] 1 : [sigma](k)|n}, (10)

introduced, and studied in [15]. The divisor minimum function

D(n) = min{k [greater than or equal to] 1 : n|d(k)} (11)

(where d(k) is the number of divisors of k) appears in [14], while the sum-of-divisors minimum and maximum functions

[epsilon] (n) = min{k [greater than or equal to] 1 : n|[sigmna](k)} (12)

[epsilon]*(n) = max{k [greater than or equal to] 1 : [sigma](k)|n} (13)

have been recently studied in [16].

For functions Q(n), [Q.sub.i](n) obtained from (3) for f (k) = k! and A = set of perfect squares, resp. A = set of squarefree numbers, see [10].

2. The aim of this note is to study some properties of the functions T(n) and T* (n) given by (1) and (2). We note that properties of T(n) in connection with "multiplicatively perfect numbers" have been introduced in [11]. For other asymptotic properties of T(n), see [7]. For divisibility properties of T(<r(n)) with T(n), see [5]. For asymptotic results of sums of type [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] see [17].

A divisor i of n is called "unitary" if (i, n/i) = 1. Let T*(n) be the product of unitary divisors of n. For similar results to [11] for T*(n), or T**(n) (i.e. the product of "bi-unitary" divisors of n), see [2]. The product of "exponential" divisors Te(n) is introduced in paper [12]. Clearly, one can introduce functions of type (1) and (2) for T(n) replaced with one of the above functions T*(n),T**,[T.sub.e](n), but these functions will be studied in another paper.

3. The following auxiliary result will be important in what follows. Lemma 1.

T (n)= [n.sup.d(n)/2], (14)

where d( n) is the number of divisors of n.

Proof. This is well-known, see e.g. [11]. Lemma 2.

T(a)\T(b), if a\b. (15)

Proof. If a|b, then for any d|a one has d|b, so T(a)\T(b). Reciprocally, if T(a)|T(b), let [[gamma].sub.p](a) be the exponent of the prime in a. Clearly, if p|a, then p|b, otherwise T(a)|T(b) is impossible. If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], then we must have [[gamma].sub.p](a) [less than or equal to] [[gamma].sub.p](b). Writing this fact for all prime divisors of a, we get a|b.

Theorem 1. If n is squarefree, then

T(n) = n. (16)

Proof. Let n = [p.sub.1][p.sub.2]... [p.sub.r], where [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] are distinct primes. The relation [p.sub.1][p.sub.2]... [p.sub.r]|T(k) gives [p.sub.i]\T(k), so there is a d|k, so that [p.sub.i]|d. But then [p.sub.i]|k for all [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], thus [p.sub.1][p.sub.2]... [p.sub.r] = n|k. Since [p.sub.1][p.sub.2]... [p.sub.k]|T([p.sub.1][p.sub.2]... [p.sub.k]), the least k is exactly [p.sub.1][p.sub.2]... [p.sub.r], proving (16).

Remark. Thus, if p is a prime, T(p) = p; if p < q are primes, then T(pq) = pq, etc.

Theorem 2. If a|b, a [not equal to] b and b is squarefree, then

T (ab) = b. (17)

Proof. If a|b, a [not equal to] b, then clearly T(b) = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] d is divisible by ab, so T(ab) [less than or equal to] b. Reciprocally, if ab|T(k), let p|b a prime divisor of b. Then p|T(k), so (see the proof of Theorem 1) p|k. But b being squarefree (i.e. a product of distinct primes), this implies b|k. The least such k is clearly k = b.

For example, T(12) = T(2 * 6) = 6, T(18) = T(3 * 6) = 6, T(20) = T(2 * 10) = 10.

Theorem 3. T(T(n)) = n for all n [greater than or equal to] 1. (18)

Proof. Let T(n)|T(k). Then by (15) one can write n|k. The least k with this property is k = n, proving relation (18).

Theorem 4. Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be distinct primes, and [a.sub.i] [greater than or equal to] 1 positive integers. Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (19)

Proof. In [13] it is proved that for A = N*, and any function f such that [F.sup.N*.sub.f]] (n) = [F.sub.f] (n) is well defined, one has

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (20)

On the other hand, if f satisfies the property

A|b f [??] f(a)|f(b)(a,b [greater than or equal to] 1), (21)

then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (22)

By Lemma 2, (21) is true for f(a) = T(a), and by using (20), (22), relation (19) follows. Theorem 5.

T([2.sup.n]) = [2.sup.[alpha], (23)

where a is the least positive integer such that

[alpha]([alpha] + 1)/2 [greater than or equal to] n. (24)

Proof. By [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] when d(k) = ([[alpha].sub.1] + 1)... ([a.sub.r] + 1). Since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], clearly [p.sub.1] = 2 and the least k is when [[alpha].sub.2] = * * * = [[alpha].sub.r] =0 and a.\ is the least positive integer with 2n < [[alpha].sub.1]([[alpha].sub.1] + 1). This proves (23), with (24).

For example, T([2.sup.2]) = 4, since [alpha] = 2, T([2.sup.3]) = 4 again, T([2.sup.4]) = 8 since [alpha] = 3, etc.

For odd prime powers, the things are more complicated. For example, for [3.sup.(n)] one has:

Theorem 6.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (25)

where ai is the least positive integer such that "l("21+1) [greater than or equal to] n, and [[alpha].sub.2] is the least positive integer such that [[alpha].sub.2]([[[alpha].sub.2] + 1) [greater than or equal to] n.

Proof. As in the proof of Theorem 5,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [p.sub.1] < [p.sub.2] < * * * < pr, so we can distinguish two cases:

a) [p.sub.1] = 2, [p.sub.2] = 3, [p.sub.3] [greater than or equal to] 5;

b) [p.sub.1] = 3, [p.sub.2] [greater than or equal to] 5.

Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] in case a), and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] in case b). So for the least k we must have [[alpha].sub.2]([[alpha].sub.1] + 1)([[alpha].sub.2] + 1) [greater than or equal to] 2n with [[alpha].sub.1] = 1 in case a), and [[alpha].sub.1]([[alpha].sub.1] + 1) [greater than or equal to] 2n in case b). Therefore [[alpha].sub.1]([[alpha].sub.1]+ 1)/2 [greater than or equal to] n and [[alpha].sub.2]([[alpha].sub.2] + 1) [greater than or equal to] n, and we must select k with the least of [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], so Theorem 6 follows.

For example, T([3.sup.2]) = 6 since for n = 2, [[alpha].sub.1] = 2, [[alpha].sub.2] = 1, and min{2 * [3.sup.1], [3.sup.2]} = 6; T([3.sup.3]) = 9 since for n = 3, [[alpha].sub.1] = 2, [[alpha].sub.2] = 2 and min{2 * [3.sup.2], [3.sup.2]} = 9.

Theorem 7. Let f: [1,[infinity]) [right arrow] [0, [infinity]) be given by f (x) = [square root (x)] log x. Then

[f.sup.-1](log n) < T(n) [less than or equal to] n, (26)

for all n [greater than or equal to] 1, where [f.sup.-1] denotes the inverse function of f.

Proof. Since n|T(n), the right side of (26) follows by definition (1) of T(n). On the other hand, by the known inequality e[member of](k) < 2[square root of (k)], and Lemma 1 (see (14)) we get T(k) < [k.sup.[square root of (k)], so log T(k) < [square root of (k)] log k = f (k). Since n|T(k) implies n [less than or equal to] T(k), so log n [less than or equal to] log T(k) < f(k), and the function f being strictly increasing and continuous, by the bijectivity of f, the left side of (26) follows.

4. The function T*(n) given by (2) differs in many aspects from T(n). The first such property is:

Theorem 8. T<(n) [less than or equal to] n for all n, with equality only if n = 1 or n = prime. Proof. If T(k)|n, then T(k) [less than or equal to] n. But T(k) [greater than or equal to] k, so k [less than or equal to] n, and the inequality follows. Let us now suppose that for n [greater than or equal to] 1, T<(n) = n. Then T(n)\n, by definition 2. On the other hand, clearly n|T( n), so T( n) = n. This is possible only when n = prime. Remark. Therefore the equality

T*(n) = n(n > 1)

is a characterization of the prime numbers.

Lemma 3. Let [p.sub.1],... ,[p.sub.r] be given distinct primes (r [greater than or equal to] 1). Then the equation

T(k) = [p.sub.1][p.sub.2] *** [p.sub.r]

is solvable if r = 1.

Proof. Since [p.sub.i]|T(k), we get [p.sub.i]|k for all i = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Thus p\... [p.sub.r]\k, and Lemma 2 implies T([p.sub.i]... [p.sub.r])|T(k) = pi... [p.sub.r]. Since [p.sub.i]... [p.sub.r]|T([p.sub.1]... [p.sub.r]), we have T([p.sub.1]... [p.sub.r]) = [p.sub.i]... [p.sub.r], which by Theorem 8 is possible only if r = 1.

Theorem 9. Let P(n) denote the greatest prime factor of n [greater than or equal to] 1. If n is squarefree, then

T*(n) = P(n). (27)

Proof. Let n = [p.sub.1][p.sub.2]... [p.sub.r], where [p.sub.1] < [p.sub.2] < *** < [p.sub.r]. If T(k)|([p.sub.1]... [p.sub.r]), then clearly T(k) [member of] {1,[p.sub.1],... ,[p.sub.r],[p.sub.1][p.sub.2],... ,[p.sub.1][p.sub.2]... [p.sub.r]}. By Lemma 3 we cannot have

T (k) [member of] {[p.sub.1][p.sub.2],... ,[p.sub.1][p.sub.2]... [p.sub.r]},

so T(k) [member of] {1,[p.sub.1],... ,[p.sub.r]}, when k [member of] {1,[p.sub.1],... ,[p.sub.r]}. The greatest k is [p.sub.r] = P(n).

Remark. Therefore T*(pq) = q for p < q. For example, T*(2 * 7) = 7, T*(3 * 5) = 5, T*(3 * 7) = 7, T*(2 * 11) = 11, etc.

Theorem 10.

T*([p.sup.n])= [p.sup.[alpha]](p = prime), (28)

where a is the greatest integer with the property

[alpha]([alpha] + 1)/2 [less than or equal to] n. (29)

Proof. If T(k)|[p.sup.n], then T(k) = [p.sup.m] for m [less than or equal to] n. Let q be a prime divisor of k. Then q = T(q)|T(k) = [2.sup.m] implies q = p, so k = pa. But then T(k) = [p.sup.[alpha]([alpha]+1)/2] with [alpha] the greatest number such that [alpha]([alpha] + 1)/2 [less than or equal to] n, which finishes the proof of (28).

For example, T*(4) = 2, since a([[alpha].sub.2]+1) [less than or equal to] 2 gives [[alpha].sub.max] = 1.

T*(16) = 4, since [alpha]([[alpha]+1)/2 [less than or equal to] 4 is satisfied with = 2.

T*(9) = 3, and T*(27) = 9 since [alpha]([[alpha]+1)/2 [less than or equal to] 3 with [[alpha].sub.max] = 2.

Theorem 11. Let p, q be distinct primes. Then

T*([p.sub.2]q) = max{p, q}. (30)

Proof. If T(k)|[p.sub.2]q, then T(k) [member of] {1,p,q,[p.sub.2],pq,[p.sub.2]q}. The equations T(k) = [p.sub.2], T(k) = pq, T(k) = [p.sub.2]q are impossible. For example, for the first equation, this can be proved as follows. By p|T(k) one has p|k, so k = [p.sup.m]. Then p([p.sup.m]) are in T(k), so m = 1. But then T(k) = p = [p.sub.2]. For the last equation, k = (pq)m and pqm([p.sup.m])(qm)(pqm) are in T(k), which is impossible.

Theorem 12. Let p, q be distinct primes. Then

T*([p.sub.3]q) = max{[p.sub.2], q}. (31)

Proof. As above, T(k) [member of] {1, p, q, pq,[p.sub.2]q,[p.sub.3]q,[p.sub.2],[p.sub.3]} and T(k) e {pq,[p.sub.2]q,[p.sub.3]q,[p.sub.2]} are impossible. But T(k) = [p.sub.3] by Lemma 1 gives kd(k) = [p.sup.6], so k = [p.sup.m], when d(k) = m + 1. This gives m( m + 1) = 6, so m = 2. Thus k = [p.sub.2]. Since p < [p.sub.2] the result follows.

Remark. The equation

T (k)= [p.sup.s] (32)

can be solved only if [k.sup.d.sup.(k)] = [p.sub.2s], so k = [p.sup.m] and we get m(m + 1) = 2s. Therefore k = [p.sup.m], with m(m + 1) = 2s, if this is solvable. If s is not a triangular number, this is impossible. Theorem 13. Let p, q be distinct primes. Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

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Jozsef Sandor

Babes-Bolyai University of Cluj, Romania

E-mail: jjsandor@hotmail.com jsandor@member.ams.org
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