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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. 














81 


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, =} ; 


Yı =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 


Nı < 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 q° || 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 
PŽ 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} : 


83 


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 
Yı <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“ Yı 














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 





84 


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. 


86