Smarandache Notions Journal, Vol. 10, No. 1-2-3, Spring 1999, pp. 82-87.

THE NORMAL BEHAVIOR OF THE SMARANDACHE FUNCTION

KEVIN FORD

Let S(n) be the smallest integer k so that n|k!. This is known as the Smarandache function and has been studied by many authors. If P(n) denotes the largest prime factor of n, it is clear that S(n) > P(n). In fact, S(n) = P(n) for most n, as noted by Erdös [E]. This means that the number, N(z), of n < x for which S(n) # P(n) is o(z). In this note we prove an asymptotic formula for N(z).

First, denote by p(u) the Dickman function, defined by

p(u)=1 (0<u<1), pw) =1- f Da (u > 1).

For u > 1 let = €(u) be defined by

It can be easily shown that

logs u E(u) = logu + loggu + O ( a ) ;

where log, z denotes the kth iterate of the logarithm function. Finally, let up = uo(x) be defined by the equation

log x = up€ (uo). The function uo(z) may also be defined directly by log xz = uo aus 1) ; It is straightforward to show that e E D)

We can now state our main result.

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Theorem 1. We have

N(a) ~ Vi(1 + log 2)

93/4 (logz log, x)9/4z} 1/40 (ug).

There is no way to write the asymptotic formula in terms of “simple” functions, but we can get a rough approximation.

Corollary 2. We have

N(x) = zexp {-(v2+ o(1))./log z logs z) .

The asymptotic formula can be made a bit simpler, without reference to the function p as follows.

Corollary 3. We have

log zr 7(1 + log2 ne eM —1 nle) ~ SUPE) tog 2) (logs 2) 2/40 exp f 6 Za,

where y = 0.5772... is the Euler-Mascheroni constant.

This will follow from Theorem 1 using the formula in Lemma 2 which relates p(u) and €(u).

The distribution of S(n) is very closely related to the distribution of the func- tion P(n). We begin with some standard estimates of the function U(z,y), which denotes the number of integers n < x with P(n) < y.

Lemma 1 {HT, Theorem 1.1]. For every e > 0,

log(u + 1) ss log z log y ~ logy’

U(z,y) = zp(u) (1 +0 (

uniformly in 1 <u < exp{(logy)9/> €}. Lemma 2 [HT, Theorem 2.1]. For u > 1,

eof) Harb f oa] = f-u (iozu + og u zio (a=) | p

Lemma 3 [HT, Corollary 2.4]. If u > 2, |v] < u/2, then

p(u)

plu v) = p(u) exp{v€(u) + O((1 + v?)/u)}. Further, if u > 1 and0 Sv <u then

plu- v) & pluje”, 82

We will show that most of the numbers counted in N(z) have

P(n) ~ exp { \/Flog tog, =} ;

=ep{i Vigri}, Ya = Yf = ep {2Viogz Toes}.

Let N, be the number of n counted by N(x) with P(n) < Yi, let No be the number of n with P(n) > Yn, and let N3 = N(x) Ni No. By Lemmas 1 and 2,

Let

< U(z, Y1) = zexp{-(1.5 + o(1)) vlog z log, z}.

For the remaining n < z counted by N(z), let p = P(n). Then either p?|n or for some prime q < p and b > 2 we have || n, q’ { p!. Since p! is divisible by q'?/¢! and b < 2logz, it follows that q > p/(3logz) > p'/?. In all cases n is divisible by the square of a prime > Y2/(3logz) and therefore

6z log z Nex y as A < zexp { -1.9 log log 2}.

¥2 Flog

Since q > p'/? it follows that q'?/4 || p!. If n is counted by N3, there is a number

b > 2 and prime q [p/b, p] so that g°|n. For each b > 2, let N3 (x) be the number of n counted in N3 such that q? || n for some prime q > p/b. We have

31 g XO Ns. KT (55) < zexp {-(5/3 + 0(1)) Vlog zlog, =} . 1

b26

Next, using Lemma 1 and the fact that p is decreasing, for 3 < b < 5 we have

sen 5, PEN E964)

Yi <p<Y2 p/bxq<p 1 log z 1 log z log p blog q <r ` oP (Zs - b) +5 ag? (epee nee ieee Yi<p<Y2 = p/2<q<p > log x <z X p "(E -@+9); Yi<p<Yo SP

By partial summation, the Prime Number Theorem, Lemma 2 and some algebra,

N3, < exp {-(15 + o(1)) vlog z log, z} :

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The bulk of the contribution to N(z) will come from N39. Using Lemma 1 we obtain

(2) won, oS) G

Yı<p<Y2 E <a<p p Ioa) 2) de a z 2) t J loga z log p log q = (1 +O ( 2) | T 5 aa a + ba A Á <p<Y2 p/2<q<p

By Lemma 3, we can write

gz— l p (WES PEP 2) = p (PEE 8) (iof o) logq logq log z The contribution in (2) from p near Y; or Y2 is negligible by previous analysis, and for fixed q [¥1, Yo/2] the Prime Number Theorem implies

Y= PE ologo *) = \$2 +0 (3),

2 Rope ase log g log p log“

Reversing the roles of p,q in the second sum in (2), we obtain

1 log x log2 /logz 7 = logs z = —? 5 _ i aR uy ( logs )) e2 P? (> (Be ) T jogp” E 3))

Yi<p<Yo2

By partial summation, the Prime Number Theorem with error term, and the change of variable u = logz/logp,

Gu Ns (1 +0 (#2) a (a) ii elu -3))a 1/udu,

where log z

ui = >z uz = bu}.

loga T’

The integrand attains its maximum value near u = uo and we next show that the most of the contribution of the integral comes from u close to up. Let

log; z ae log, £

iit w,= Kuo, w= w (

where K is a large absolute constant. Let J; be the contribution to the integral in (3) with |u uo| > wo, let Iz be the contribution from w < |u ug| < wo, let I3 be the contribution from wz < |u up| < wi, and let I4 be the contribution from ju uo] < we. First, by Lemma 2, the integrand in (3) is

[Cipra vgs}. (s

log z

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The function 1/c + c/2 has a minimum of V2 at c = V2, so it follows that

ia exp {- (v2+10 *) Vlogs log, =}.

Let u = up v. For w, <S |v| < wo, Lemma 2 and the definition (1) of up imply that the integrand in (3) is

l < pluo) exp { velu) = T (+2 L 2 oe =) +o(Ż + log) up F oe y2 < p(uo)z ° exp Ea )} j2 & p(uo)z 1/* exp {70.955 log z) for K large enough. It follows that

In K upp(uo)x 1/*° exp{—20 logy 1} « (logz) 1 p(uo)z 1°.

For the remaining u, we first apply Lemma 3 with v = 2 and v = 3 to obtain

Ugtwi 2€(u) los 2 Ig3+ig={1+0O \/ 1082 = ) p(u)x t/g = + 06 Su) du Og T a u ogr

o wi

We will show that J3 + I4 > p(uo)z 1 (log z)3/?, which implies

uo+twi 2€(u) (4) N(z)= (1+0/ REE) ) i: p(u)z me poe eca) du.

ipui u + gz

This provides an asymptotic formula for N(x), but we can simplify the expression somewhat at the expense of weakening the error term. First, we use the formula

_ logs u E(u) = logu + log, u + O (7) :

and then use u = uo + O(u Ae and (1) to obtain

Ni

I+ = (1+0 (282 ayes (1 + log 2)z(log x)? (log, z)

log, T

uotwi / p(ujx 1! du.

o Wi

By Lemma 3, when w < |u| < wi, where u = up v, we have

as L logz fu v v pluo —v)r = ® < p(ug)x *o exp < vE(uo) (2+

i & pluo) “0 exp

( < p(uo)z 70 exp -2 log z) {

< pluo)z 70 (logz z) 3

provided K is large enough. This gives i plu) /* du < p(uo)e ™*°(logz)!/4(logy x) 3. w2K|u uo[<wi

For the remaining v, Lemma 3 gives

pluo —v)z Vo ») = (a +O (2822 )) p(uo)x 1/™ exp f- tog}

Therefore, uptw2 w2 21 pluo) lgs f plu)z "du = (a +0 (22) / epf- ae} dv. uo w2 w2 0

The extension of the limits of integration to (—oo, c0) introduces another factor 1+ O((logz) 1), so we obtain

seve m(1 + log 2) EN Btt = (1+0 (BEE)) aa oezo a) aue \$

and Theorem 1 follows. Corollary 2 follows immediately from Theorem 1 and (1). To obtain Corollary 3, we first observe that é’(u) ~ u | and next use Lemma 2 to

write 2 i: €(t) dt pluo) Jorg exp i E( \$ By the definitions of and ug we then obtain uo E(uo) v J a= | wise Za 1 0 co = ef(uo) _ vb o

log z log z = a is!

Uo

Corollary 3 now follows from (1).

REFERENCES

[E] P. Erdős, Problem 6674, Amer. Math. Monthly 98 (1991), 965. [HT] A. Hildebrandt and G. Tenenbaum, Integers without large prime factors, J. Théor. Nombres Bordeaux 5 (1993), 411-484.

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