Presenting an application to a diophantine equation. The application is not of the theorem per se, but rather of the counting method used to prove the theorem

Luca, Florian

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THE AVERAGE SMARANDACHE FUNCTION Florian Luca Mathematical Institute, Czech Academy of Sciences Zitrui 25, 115 67 Praha 1 Czech Republic For every positive integer n let S(n) be the minimal positive integer m such that n I mL For any positive number x ~ 1 let 1 A(x) = - 2: Sen) x n~'" (1) be the average value of S on the interval [1, xl. In [6], the authors show that (2) where Cl can be made rather small provided that x is enough large (for example, one can take Cl = .215 and C2 = 45.15 provided that x > 1470). It is interesting to mention that by using the method outlined in [6]. one gets smaller and smaller values of Cl for which (2) holds provided that x is large, but at the cost of increasing C2! In the same paper, the authors ask whether it can be shown that 2x A(x) < -I -ogx and conjecture that, in fact, the stronger version x A(x)<-l -ogx (3) (4) might hold (the authors of [6] claim that (4) has been tested by Ibstedt in the range x S 5 . 106 in [4J. Although I have read [4J carefully, I found no trace of the aforementioned computation!). In this note, we show that -I x is indeed the correct order of magnitude of ogx A(x). For any positive real number x let 7I"(x) be the number of prime numbers less then or equal to x, B(x) = xA(x) = 2: Sen), l~n~'" 1 E(x) = 2.510glog(x) + 6.2 + -. x We have the following result: Theorem . . 5(7I"(x) - r.(.jX» < A(x) < 7I"(x) + E(x) for all x ~ 3. Inequalities (7), combined with the prime number theorem, assert that 19 (5) (6) (7) .5::; liminf A~x) ::; limsup A~x) ::; 1, "-co __ .,-co __ log x log x which says that -1 x is indeed the right order of magnitude of A(x). The natural ~x . conjecture is that, in fact, Since A(x)=-lx +0(-;-). ogx log x X( 1) X( 3) -- 1+-- <1I'X <-- 1+--log x 2logx () logx 210gx (8) for x ~ 59, it follows, by our theorem, that the upper bound on A(x) is indeed of the type (8). Unfortunately, we have not succeeded in finding a lower bound of the type (8) for A(x). The Proof We begin with the following observation: Lemma. Suppose that n = pr1 ".p~' is the decomposition of n in prime factors (we assume that the Pi'S are distinct but not necessarily ordered). Then: L S(n)::; maxf=l(O'iPi)' (9) 2. Assume that O'lPl = ma.xf=l(O'iPi). If 0'1::; PI, then Sen) = O'lPl· 3. Sen) > O'i(Pi - 1) for all i = 1, . __ , k. Proof. (10) For every prime number P and positive integer k let ep(k) be the exponent at which P appears in k!. L Let m ~ ma.xf=l(O'iPi). Then ep.(m)=2:l~J ~ l~j ~O'i fori=l, ... , k . • ~l P. P. This obviously implies n I m!, hence m ~ Sen). 2. Assume that 0'1 ::; Pl. In this case, S(n) ~ O'lPl. By 1 above, it follows that in fact Sen) = O'lPl. 3. Let m = Sen). The asserted inequality follows from O'i ::; ep • (m) = 2: l n: j < m f 1~ = . r: 1 . • ~l P. .~l P. P. 20 The Proof of the Theorem. In what follows p denotes a prime. We assume 'Z > 1. The idea behind the proof is to find good bounds on the expression R(x) - R(-./X) = L Sen). fo<n'5 z Consider the following three subsets of the interval I = (y'X, xl: C1 = {n E I I S(n) "is not a prime}, C2 = {n E I I Sen) = p ~ y'X}, C3 = {n E I I S(n) = p > y'X}. (11) Certainly, the three subsets above are, in general, not disjoint but their union covers I. Let D;(x) = L Sen) for i = 1, 2, 3. nEG, Clearly, max(D;(x) Ii = 1, 2, 3) ~ R(x) - R(y'X) ~ D1(x) + D2(x) + D3(X), (12) We now bound each D; separately. The bound for D1 . Assume that m E C1 . By the Lemma, it follows that S( m) ~ Otp for some pO II m and Ot > L First of all, notice that SCm) ~ Ot...;m. Indeed, this follows from the fact that for Ot 2: 2. In particular, from the above inequality it follows that p ~ ...;m ~ y'X. Write now m = pOk. Since m ~ x, it follows that k ~ x/po, These considerations show that In the above formula (13), we used the fact that ~ OtZo-1 - ~(_1_) -1- (_1_)2 -1- 2z - z2 for Izi < 1 L.. - dz 1 - z - 1 - z - (1 - z)2 0;::2 with z = lip. Since it follows that -;-2..:..p_--:-:l~ < ~ (p -1)2 - 4p 21 for p 2: 3, (13) From a formula from [5], we know that 1 L - < loglogy+ 1.27 p~y P for all y > 1. . Hence, inequality (14) implies D1(x) < x( 2.375 + 1.25 (log logv'Z + 1.27)) < x (3.1 + 1.25 log log x). (15) The bound for D2 Assume that SCm) = p. Then m = py where p does not divide y. Since m > .;x, it follows that .;x x -<y~-p p Since p ~ Vx, it follows that at least one integer in the above interval is a multiple of p; hence, cannot be an acceptable value for y. This shows that there are at most l x - .;xJ x - .;x --- <---P - P possible values for y. Hence, (16) Bounds for D3 Assume SCm) = p for some p > .;x. Then, m = py for some y < x/p. Hence, (17) Notice that, unlike in the previous cases, (17) is in fact an equality. Since z 2:: l z J > .5z for all real numbers z> I, it follows, from formula (17), that Denote now by F(x) = 3.1 + 1.25 log log(x) From inequalities (12), (15), (16) and (17), it follows that (18) .5x(1I'(x) - 7l'(-/X» < D3(X) < B(x) - B(v'Z) < D1(x) + D2(x) + D3(X) < xF(x) + (x - VZ)7l'( v'Z) + x( 1I'(x) - 11'( v'Z» = X7r(X) - -/X1I'( -/X) + xF(x). (19) The left inequality (7) is now obvious since B(x) > B(v'Z) + .5x(1I'(x) -1I'(.;x» 2:: 1 + .5x(1I'(x) - 7l'(.;x). 22 For the right inequality (7). let G(x) = n(x). Formula (19) can be rewritten as B(x) - B(v'X) < G(x) - G(v'X) + xF(x). (20) Applying inequality (20) with x replaced by .../i, x1/4, ... , x i / 2 ' until x 1/ 2' < 2 and summing up all these inequalities one gets • B(x) - B(l) < G(x) + L>1/2iF(x1/2i). (21) .=0 The function F(x) is obviously increasing. 'Hence, • B(x) < 1 + G(x) + F(x) I>1/2'. (22) ;=0 To finish the argument, we show that • x ~ 1:xl/2i. (23) ;=1 Proceed by induction on s. If s = 0, there is nothing to prove. If s = 1, this just says that x > .;x which is obvious. Finally, if s ~ 2, it follows that x ~ 4. In particular, x ~ 2.../i or x - .../i ~ v'X. Rewriting inequality (23) as which is precisely inequality (23) for .../i. This completes the induction step. Via inequality (23), inequality (22) implies or B(x) < l+n(x)+2xF(x)= l+X7r(x)+2x(3.l+1.251oglogx) (24) 1 A(x) < IT(X) + - + 6.2 + 2.5loglogx = IT(X) + E(x). x Applications From the theorem, it follows easily that for every ! > 0 there exists Xo such that A(x) < (1 + !)-l x . ogx (25) In practice, finding a lower bound on Xo for a given f, one simply uses the theorem and the estimate x ( 3) lTx<--1+--() log x 2 log x for x > 1. (26) 23 (see [5]). By (7) and (26), it now follows that (25) is satisfied provided that x 1 (3 ) - > - --+E(x) . log x f 210g2 X For example, when f = 1, one gets for f = .5, one gets x A(x) < 2-1 -ogx z A(z) < 1.5-1 -ogx and for f = 0.1 one gets z A(x) < 1.1-1 -ogx for x ~ 64, for z ~ 254 for x ~ 3298109. (27) (28) (29) Of course, inequalites (27)-(29) may hold even below the smallest values shown above but this needs to be checked computationally. In the same spirit, by using the theorem and the estimation x ( 1) 11" X >--1+--() log x 210gx for x ~ 59 (see [5]) one can compute, for any given (, an initial value Xo such that x A(x) > (.5 - f)-l -ogx For example, when (= 1/6 one gets 1 x A(x) >--3 log x forx>xo. for x ~ 59. (30) Inequality (30) above is better than the inequality appearing on page 62 in [2J which asserts that for every Q > 0 there exists Xo such that A(x) > xOi/x for x > Xo (31 ) because the right side of (31) is bounded and the right side of (30) isn't! A diophantine equation In this section we present an application to a diophantine equation. The ap-plication is not of the theorem per se, but rather of the counting method used to prove the theorem. Since S is defined in terms of factorials, it seems natural to ask how often the product S(l) . S(2) . ' ... Sen) happens to be a factorial. Proposition. The only solutions of S(l) . S(2) .. ., . Sen) = m! (32) 24 are given by n = m E {I, 2, ... , 5}. Proof. We show that the given equation has no solutions for n ? 50. Assume that this is not so. Let P be the largest prime number smaller than n. By Tchebysheff's theorem, we know that P ? n/2. Since S(P) = P, it follows that P I m!. In particular, P :5 m. Hence, m ? n/2. We now compute an upper bound for the order of 2 in S(I) . S(2) ..... Sen). Fix some /3 ? 1 and assume that 1: is such that 2(3 II S(1:). Since S(k) = max(S(pCk) I pck II k), it follows that 2(3 /I S(pa) for some pa II k. We distinguish two situations: Casel. p is odd. In this case, 2(3p I S(pa). If /3 = 1, then Q = 2. If /3 = 2, then Q = 4. For /3 ? 3, one can easily check that Q ? 2(3 - /3 + 1 (indeed, if a :5 2(3 - /3, then one can check that pa I (2(jp - 1)! which contradicts the definition of S). In particular, p2~-rJ+l I k. Since 2,,-1 ? x + 1 for x ? 3, it follows that Q ? 2/3-1 + 2. Since k :5 n, the above arguments show that there are at most n for /3 = 1, 2 p2~ and n for /3? 3 p2~-1+2 integers k in the interval [1, n] for which p I k, S(1:) = S(pa), where Q is such that pa II k and 2/3 1/ S(k). Case 2. p = 2. If /3 = 1, then k = 2. If /3 = 2, then k = 4. Assume now that /3 ? 3. By an argument similar to the one employed at Case 1, one gets in this case that a ? 2(3 - /3. Since 2a /I k, it follows that 22~-(3 I k. Since k :5 n, it follows that there are at most n such k's. From the above anaysis, it follows that the order at which 2 divides S( 1) . S(2) . ... . Sen) is at most ( 1 2 /3) /3 e2 < 3 + n L "2 + 4" + L 2~-1+2 + n L 2 2'-/3 . ,~.. P P (3)3 P (3)3 , odd. - -(the number 3 in the above formula counts the contributions of 5(2) S( 4) = 4). We now bound each one of the two sums above. For fixed p, one has (38) 2 and 1 2 /3 1 2 3 4 'Y p2 p2 + p4 + L p2fJ-1+2 = p2 + p4 + p6 + plO + ... < L p2.., = (p2 _ 1)2· (39) (j~3 ..,~l 25 Hence, ~(1 2 f3) p2 6 2" +"4 + E p2~-1+2 < E (P2 -1)2 < .245 ~. p p ~~ p~d ,odd -We now bound the second sum: ~f3 345 3 ~ f3 6 22~-~ = 25 + 212 + 227 + ... < 26 + 6 22+4(P-2) = P~3 P~3 3 1 (~'Y + 2) 3. 1 (15 31 ) 26 +:4 6 16-' = 26 + 4 16 + 225 < .099 -,>1 From inequalities (38), (40) and (41), it follows that e2 < 3 + .344n. (40) ( 41) (42) We now compute a lower bound for e2. Since e2 = e2(m!), it follows, from Lemme 1 in [1] and from the fact that m ~ nf2, that log(m + 1) n log(nj2 + 1) e2 > m - > - - --=~:........:~..;. - log 2 - 2 log 2 . (43) From inequalities {42) and (43), it follows that 3 344 5 log(.5n + 1) +. n>.n- 12 ' - og which gives n :5 50. One can now compute 5(1) ·5(2) . '" . 5(n) for all n :5 50 to conclude that the only instances when these products are factorials are n = 1, 2, ... , 5. We conclude suggesting the following problem: Problem. Find all positive integers n such that 5(1), 5(2), ... , 5(n2 ) can be arranged in a latin square. The above problem appeared as Problem 24 in SNJ 9, (1994) but the range of solutions was restricted to {2, 3, 4, 5, 7, 8, 10}. The published solution was based on the simple observation that the sum of all entries in an n x n latin square has to be a multiple of n. By computing the sums B(x2 ) for x in the above range, one concluded that B(x2 ) =t 0 (mod x) which meant that there is no solution for such x'ses. It is unlikely that this argument can be extended to cover the general case. One should notice that from our theorem, it follows that if a solution exists for some n > 1, then the size of the common sums of all entries belonging to the same row (or column) is :::! mr(n2 ). Addendum After this paper was written, it was pointed out to us by an annonymous referee that Finch [3} proved recently a much stronger statement, namely that . log(x) 11'2 hm --' A(x) = - = 0.82246703 ... "-00 x 12 (44) 26 Finch's result is better than our result which only shows that the limsup of the expression log(z)A(z)jz when z goes to infinity is in the interval [0.5, 1]. References [1] Y. Bugeaud & M. Laurent, "Minoration effective de la distance p-adique entre puissances de nombres algebriques", J. Number Theory 61 (1996), pp. 311-342. [2] C. Dumitrescu & V. Seleacu, "The Smarandache Function", Erhus U. Press, 1996. [3] S.R. Finch, "Moments of the Smarandache Function", SNJ 11, No. 1-2-3 (2000), p. 140-142. [4] H. Ibsted, "Surfing on the ocean of numbers", Erhus U. Press, 1997. [5J J. B. Rosser & L. Schoenfeld, «Approximate formulas for some functions of prime numbers", minois J. of Math. 6 (1962), pp. 64-94. [6] S. Tabirca & T. Tabirca, "Two functions in number theory and some upper bounds for the Smarandache's function", SNJ 9 No. 1-2, (1998), pp. 82-91. 1991 AMS Subject Classification: llA25, llL20, llL26. 27 

Minoration effective de la distance p-adique entre puissances de nombres algebriques&quot;,

Moments of the Smarandache Function&quot;,

Surfing on the ocean of numbers&quot;, Erhus U.

The Smarandache Function&quot;, Erhus U.

Two functions in number theory and some upper bounds for the Smarandache's function&quot;,

THE AVERAGE SMARANDACHE FUNCTION

https://zenodo.org/record/9197/files/TheAverSF.pdf

THE AVERAGE SMARANDACHE FUNCTION

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