This papers deals with the introduction and preliminary study of the Smarandache minimum and maximum functions

Sandor, J.

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Scientia MagnaVol. 1 (2005), No. 2, 162-166The Smarandache minimum and maximumfunctionsJo´zsef Sa´ndorBabes-Bolyai University of Cluj, RomaniaAbstract This papers deals with the introduction and preliminary study of the Smarandache minimumand maximum functions.Keywords Smarandache minimum and maximum functions; arithmetical properties.1. Let f : N∗ −→ N be a given arithmetic function and A ⊂ N a given set. The arithmeticfunctionFAf (n) = min{k ∈ A : n | f(k)} (1)has been introduced in [4] and [5].For A = N, f(k) = k! one obtains the Smarandache function; For A = N∗, A = p ={2, 3, 5, ·} = set of all primes, one obtains a functionP (n) = min{k ∈ P : n | k!} (2)For the properties of this function, see [4] and [5]. The “dual” function of (1) has beendefined byGAg (n) = max{k ∈ A : g(k) | n}, (3)where g : N∗ −→ N is a given function, and A ∈ N is a given set. Particularly, forA =N∗, g(k) = k!, one obtains the dual of the Smarandache function,S∗(n) = max{k ≥ 1 : k! | n} (4)For the properties of this function, see [4] and [5]. F.Luca [3], K.Atanassov [1] and L.le [2]have proved in the affirmative a conjecture of the author.For A = N∗ and f(k) = g(k) = ϕ(k) in (1), resp.(3) one obtains the Euler minimum, resp.maximum-function, defined byE(n) = min{k ≥ 1 : n | ϕ(k)}, (5)Vol. 1 The Smarandache minimum and maximum functions 163E∗(n) = max{k ≥ 1 : ϕ(k) | n} (6)For the properties of these function, see [6]. When A = N∗, f(k) = d(k) =number of divisorsof k, one obtains the divisor minimum function (see [4], [5] and [7])D(n) = min{k ≥ 1 : n | d(k)}. (7)It is interesting to note that the divisor maximum function (i.e., the “ dual” of D(n)) given byD∗(n) = max{k ≥ 1 : d(k) | n} (8)is not well defined! Indeed, for any prime p one has d(pn−1) = n | n and pn−1 is unbounded asp −→∞. For a finite set A, however DA∗ (n) does exist. On one hand, it has been shown in [4]and [5] that∑(n) = min{k ≥ 1 : n | σ(k)} (9)(denoted there by Fσ(n)) is well defined. (Here σ(k) denotes the sum of all divisors of k). Thedual of the sum-of-divisors minimum function is∑∗(n) = max(k ≥ 1 : σ(k) | n}) (10)Since σ(1) = 1 | n and σ(k) ≥ k, clearly ∑ ∗(n) ≤ n, so this function is well defined (see [8]).2. The Smarandache minimum function will be defined for A = N∗, f(k) = S(k) in (1).Let us denote this function by Smin :Smin(n) = min{k ≥ 1 : n | S(k)} (11)Let us assume that S(1) = 1, i. e., S(n) is defined by (1) for A = N∗, f(k) = k! :S(n) = min{k ≥ 1 : n | k!} (12)Otherwise (i.e.when S(1) = 0) by n | 0 for all n, by (11) for one gets the trivial functionSmin(n) = 0. By this assumption, however, one obtains a very interesting (and difficult)function smin given by (11). Since n | S(n!) = n, this function is correctly defined.The Smarandache maximum function will be defined as the dual of Smin :Smax(n) = max(k ≥ 1 : S(k) | n}. (13)We prove that this is well defined. Indeed, for a fixed n, there are a finite number ofdivisors of n, let i | n be one of them. The equationS(k) = i (14)is well-known to have a number of d(i!) − d((i − 1)!) solutions, i. e., in a finite number. Thisimplies that for a given n there are at most finitely many k with S(k) | k, so the maximum in(13) is attained.164 Jo´zsef Sa´ndor No. 2Clearly Smin(1) = 1, Smin(2) = 2, Smin(3) = 3, Smin(4) = 4, Smin(5) = 5, Smin(6) =9, Smin(7) = 7, Smin(8) = 32, Smin(9) = 27, Smin(10) = 25, Smin(11) = 11, etc, which can bedetermined from a table of Smarandache numbers:n 1 2 3 4 5 6 7 8 9 10 11 12 13S(n) 1 2 3 4 5 3 7 4 6 5 11 4 13n 14 15 16 17 18 19 20 21 22 23 24 25S(n) 7 5 6 7 6 19 5 7 11 23 4 10We first prove that:Theorem 1. Smin(n) ≥ n for all n ≥ 1, with equality only forn = 1, 4, p(p = prime) (15)Proof. Let n | S(k). If we would have k < n, then since S(k) ≤ k < n we should getS(k) < n, in contradiction with n | S(k). Thus k ≥ n, and taking minimum, the inequalityfollows. There is equality for n = 1 and n = 4. Let now n > 4. If n = p =prime, thenp | S(p) = p, but for k < p, p †S(k). Indeed, by S(k) ≤ k < p this is impossible. Reciprocally, ifmin{k ≥ 1 : n | S(k)} = n, then n | S(n), and by S(n) ≤ n this is possible only when S(n) = n,i. e., when n = 1, 4, p(p = prime).Theorem 2. For all n ≥ 1,Smin(n) ≤ n! ≤ Smax(n) (16)Proof. Since S(n!)=n, definition (11) gives the left side of (16), while definition (13)gives the right side inequality.Corollary. The series∑n≥11Smin(n)is divergent, while the series∑n≥11Smax(n)is convergent.Proof. Since∑n≥11Smax(n)≤∑n≥11n!= e − 1 by (16), this series is convergent. On theother hand, ∑n≥11Smin(n)≥∑p1Smin(p)=∑p1p= +∞,so the first series is divergent.Theorem 3. For all primes p one hasSmax(p) = p! (17)Proof. Let S(k) | p. Then S(k) = 1 or S(k) = p. We prove that if S(k) = p, then k ≤ p!.Indeed, this follows from the definition (12), since S(k) = min{m ≥ 1 : k | m!} = p impliesk | p!, so k ≤ p!. Therefore the greatest value of k is k = p!, when S(k) = p | p. This provesrelation (17).Theorem 4. For all primes p,Smin(2p) ≤ p2 ≤ Smax(2p) (18)Vol. 1 The Smarandache minimum and maximum functions 165and more generally; for all m ≤ p,Smin(mp) ≤ pm ≤ Smax(mp) (19)Proof. (19) follows by the known relation S(pm) = mp if m ≤ p and the definition (11),(13). Particularly, for m = 2, (19) reduces to (18). For m = p, (19) givesSmin(p2) ≤ pp ≤ Smax(p2) (20)This case when m is also an arbitrary prime is given in.Theorem 5. For all odd primes p and q, p < q one hasSmin(pq) ≤ qp ≤ pq ≤ Smax(pq) (21)(21) holds also when p = 2 and q ≥ 5.Proof. Since S(qp) = pq and S(pq) = qp for primes p and q, the extreme inequalitiesof (21) follow from the definition (11) and (13). For the inequality qp < pq remark that this isequivalent to f(p) > f(q), where f(x) = ln xx (x ≥ 3).Since f ′(x) = 1−ln xx2 = 0 ⇔ x = e immediately follows that f is strictly decreasing forx ≥ e = 2.71· From the graph of this function, since ln 22 = ln 44 we get thatln 22<ln 33,butln 22>ln qqfor q ≥ 5. Therefore (21) holds when p = 2 and q ≥ 5. Indeed, f(q) ≤ f(5) < f(4) = f(2).Remark. For all primes p, qSmin(pq) ≤ min{pq, qp} (22)andSmax(pq) ≥ max{pq, qp}. (23)For p = q this implies relation (21).Proof. Since S(qp) = S(pq) = pq, one hasSmin(pq) ≤ pq, Smin(pq) ≤ qp, Smax(pq) ≤ pq, Smax(pq) ≤ qp.166 Jo´zsef Sa´ndor No. 2References[1] K.Atanassov, Remark on Jozsef Sandor and Florian Luca’s theorem, C. R. Acad. Bulg.Sci., 55(10)(2002), 9-14.[2] M.Le, A conjecture concerning the Smarandache dual function, Smarandache NotionsJ., 14(2004), 153-155.[3] F.Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci., 53(6)(2000),35-38.[4] J.Sandor, On certain generalizations of the Smarandache function, Notes Number The-orem Disci. Math., 5(2)(1999),41-51.[5] J.Sandor, On certain generalizations of the Smarandache function, Smarandache No-tions Journal, 11(2000), 202-212.[6] J.Sandor, On the Euler minimum and maximum functions(to appear).

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