Charles Ashbacher

In this paper we shall investigate some aspects involving

Smarandache function, S:N'-->N, s(n) = min {m | n divide m!}.

1. THE MINIMUM OF S(n)/n Which is minimumum of S(n)/n if n > 1? 1.1. THEOREM: a) S(n)/n has no minimum for n > 1. _b) lim S(n)/n as n goes to infinity does not exist. Proof:

a) Since S(n) > 1 for nol it follows that S(n)/n > 0. Assume that S(n)/n has a minimum and let the rational fraction be represented by r/s. By the infinitude of the natural numbers, we can find a number m such 2/m < r/s. Using the infinitude of the primes, we can find a prime number p > m. Therefore, we have the


27D < 27m L T/S


We have S(p-p) = S(p) = 2p. It is known that S(p:p)s2p. The

ratio of S(p°)/(p-p) is then

2p/ (p?) = 2/p

And this ratio is less than r/s, contradicting the assumption of the minimum.

b) Suppose lim S(n)/n exists and has value r. Now choose, e > 0 and e < 1/p where p is a twenty digit prime. Since Sip) = p, S(p)/p = 1.

However, S(p-p) = 2p, so the ratio S(n)/n = 2p/(p-p) = 2/p. Since

p is a twenty digit prime,

| S(p)/p - S(p'p)/ (pp) | > e by choice of e.

so the limit does not exist.

2. THE DECIMAL NUMBER WHOSE DIGITS ARE THE VALUES OF SMARANDACHE FUNCTION IS IRRATIONAL. Unsolved problem number (8) in [1] is as follows: Is r = 0,0234537465114..., where the sequence of digits is S(n), n 21, an irrational number? The number r is indeed irrational and this claim will be proven below.

The following well-known results will be used. DIRICHLET’S THEOREM:

If d > 1 and a # 0 are integers that are relativey prime, then

the arithmetic progression


A, ‘ay. +d; a+ 2d, a t 3075-2

contains infinitely many primes.

Proof of claim: Assume that r as defined above is rational. Then after some m digits, there must exist a series of digits tju, “Cs, Cin eae Enr

such that

ye 0023453746114 aabt Ott tea En

where s is the m-th digit in the decimal expansion. Now, construct the repunit number consisting of 10n 1's. a = 11111 ... 111 10n times and let d = 1000 ...`00 10n + 1 0’s Since the only prime factors of d are 2 and 5, it is clear that a and d are relatively prime and by Dirichiet’s Theorem, the sequence a, a + d, a+ 2d, ... must contain primes. Given the number of 1's in a and the fact that S(p) =p, it follows that the sequence of repeated digits in r must consist entirely of i’s.

Now, construct the repdigit number constructed from 10n 3's

a = 2333-232

10n times


and using d = 10000...00 10n + 1 0's we again have a and d relatyvely prime. Arguments Similar to those used before forces the conclusion that the sequence of repeted digits must consist entirely of 3’s. This is of course impossibile and therefore the assumption of

rationality must be false.



The following problem is listed as unsolved problem number (7)

in [1]

Are the points p(n) =S(n)/n uniformly distributed in the

interval (0,1)? The answer is no, the interval (0.5,1.0) contains only a finite number of points p(n).

3.1. LEMMA:

Sip) , sen p* p**t

For p prime and k>0.


Proof: It ig well-known that S(p*)=j:p where jsk

Therefore, forming the expressions

Á ee

where m must have one of the two values {j, j+t} With the restrictions on the values of m and p, it is clear that ad Oo p which implies that Sip*) Sips?) p* pr which is the desired result. Equality occurs only when p=2, j=1 and


3.2. LEMMA:

The interval (0.5,1.0) contains only a finite number of points

p(n), where

p(n) = 312) and n is a power of a prime. Proof: If n=p S(p) 24 , outside the interval.

pP Start with the smallest prime p=2 and move up the powers of 2

S(2:2) L4 23 (2'2) 4

5 (2°22)

ae (2'22) 8


S$(2°2:2°2) 6 = Si2222) 2 6 <0.5 | (2222) 16 R

And applying the previous lemma, all additional powers of 2

will yield a value less than 0.5.

Taking the next smallest prime p=3 and moving up the powers

of 3 S(33) _ 6 (3:3) 9 S(333) _ 9 22i ee Gaa a7

and by the previous lemma, all additional powers of 3 also yield a value less than 0.5. Now, if p>3 and p is prime

SPP) -20.5 (pp) P l so all other powers of primes yield values less than 0.5 and we are

done. 3.3. THEOREM: The interval (0.5,1.0) contare only a finite number of points p(n) where p(n) = Sa Proof: It is well-known that if


N=D. p: p”. nEeD a then S(n) =max{S(P} ))} Applying the well-known result with the formula for p(n)

S(p“ m ee Ł

Di [| 2: ze:

which 1s clearly less than


Theorefore, applying Lemma 2, we get the desired results.

3.4. COROLLARY: The points p(n)=S(n)/n are not evenly distributed in the

interval (0,1).



Unsolved problem number 31 in [1] is as follows. Does the Smarandache function veryfy a Lipschitz condition? In

other words, is there a real number L such that

| S(m) - S(n) |s L|[m-n| for all m,n i 107152730245)

4.1. THEOREM The Smarandache function does not verify a Lipschitz cöndıition:


Suppose that Smarandache function does indeed satisfy a

Lipschitz condition and let L be the Lipschitz constant.

Since the numbers of primes is infinite, is possible to fiind

a prime p such that



Now, examine the numbers (p-1) and (p+1). Clearly, at least one must not be a power of two, so we choose that one call it m. Factoring m into the product of all primes equal to 2 and everything else, we have m= 25n

Then S(m) = max {$(2*),S(n)} and because $(2*) < 2* . we have

m Sim) < 5 And so, Isip) - S(m)|> |p - =| ee Since |p-m|=21 by choice of m, we have a violation of the Lipschitz condition, rendering our original assumption false. Therefore, the Smarandache function does not Satisfy a

Lipschitz condition.


One of the unsolved problems in [1] involves a relationship between the Smarandache and factorial functions.

Solve the Diophantine Equation S(m) = n!

where m and n are positive integers. This equation is always solvable and the number of solutions is a function of the number of primes less than or equal to n.

5.1. LEMMA: Let be a prime. Then the range of the sequence


S(P SDH); S (DD D)pri

will contain all positive integral multiples of p. Proof: It has already been proven [2] that for all integers

k > 0, there exists another integer m > 0, such that

S(p*)k = mp where msk

and in particular

S(p) =p

So the only remaining element of the proof is to show that m takes on all possible integral values greater than 0.

Let p be an arbitrary prime number and define the set M = { all positive integers n such that there is no positive integer k such that S(p*) = mp } | and assume that M is not empty.

Since M is non-empty subset of the natural numbers, it must have a least element. Call that least element m. It is clear that aa ck.

Now, let j be the largest integer such that S(p) = (n= 1) -p

and consider the exponent j + 1. By the choise of j, it follows that either 1) S(p7*t) = mp


2) §{pi**) np where n> m

in the first case, we have a contradiction of our choise of m,


so we proceed to case (2).

However, it is a direct consequence of the definition of prime numbers that if ((m - 1)-p)! contains j instances of the prime p, then m-p is the smallest number such that (m-p)! contains more than j instances of p. Then, using the definition of Smarandache function where we choose the smallest number having the required number of instances we have a contradiction of case (2).

Therefore, it follows that there can be no least element of the set M, so M must be empty.

5.2.THEOREM: Let n be any integer and p a prime less than or

equal to n. Then, there is some integer k such that

S(p*) =n!

Therefore, each equation of the form S(m) = n! has at least “’p solutions, where “p is the number of primes less than or equal ton.


Since n! is an integral multiple of p for p any prime less than or equal to n, this is a direct consequence of the lemma.

Now that the question is known to have multiple solutions, the next logical question is to determine how many solutions there are.

5.3. DEFINITION: Let NSF(n) be the number of integers m, such

that S(m) = n!.

From the fact hat s(n) = max {S(p,‘)} , we have the following obvious result.



Let n be a positive integer, q a prime less than or equal to

n and k another positive integer such that s(q*) =n! . Then, all

numbers having the prime factorization form m = g*p.*p,"p,?..-D.*

where §(q*) > $(p;‘) will also be solutions the equation S(m) = n!

To proceed further, we need the following two obvious lemmas. 5.4. LEMMA: If p is a prime and m and n nonnegative integers m > n, then S(p") s S(p”). 5.5. LEMA: If p and q are primes such that p < q and k > 0, then S(p“) < S(q*). The following theorem gives an initiai indication regarding how fast NSF(n) grows as n does. 5.6. THEOREM: Let q be a prime number and k an exponent such that S(q) = n! Let p,,P),.--,P, be the list of primes less than q. Then the number of solutions to the equation S(m) = n! where m contains

exactly k instances of the prime q is at least (k +1)". Proof: Applyng the two lemmas, the numbers m = p,'p,’p,°...p,’g*

where all of exponents on the primes p; are at most solutions to the equation. Since each prime pi can have (k + 1), {0,1,2,...,k} different values for the exponent, simple counting gives the result.

Since this procedure can be repeated for each prime less than or equal to n, we have an initial number of solutions given by the



> DE a en ž-2 where s is the number of primes less then or equal to n, k is the

integer such that

S(p;*) =n! And even this is a very poor lower bound on the number of

solutions for n having any size.

5.7. COROLARY: Let q be a prime such that for some k S(q‘) = n!. Then if p is any prime such that there is some integer j such that S(p) < S(qt), then the product of any solution and p any power less than or equal to j will also be a solution.

Proof: Clear.

If q is the largest prime less than or equal ton, it is easy to show for "large" n that there are primes p > n > q that satisfy the above conditions. If any prime, then by Bertrand’s Postulate, another prime r can be found in the interval p > r > 2p. Since q < n < 2n < n! for n > 2 and S(p) = p, we have one such prime. Expanding this reasoning, it folows that the number of such primes is at least j, where j is the largest exponent of 2 such that q2 s n!, or put another way, the largest power of 2 that 1s less than or equal to n!/q.

Since there are so many solutions to the equation S(m) = n!, it is logical to consider the order of growth of the number of solutions rather than the actual number.

It is well known that the number of primes less than or equal to n is asymptotic to the ratio n/in(n). Now, let p be the largest

prime less than n. As n gets larger, it is clear that the factor m


such that mp = n! grows on the order of a factorial. Since ms k, where k is the exponent on the power p, it follows that the number grows on the order of the product of factorials. Since the number of items in the product depends on the number of primes q such that q < mp =n!, it follows that this number also grows on the order of a factorial.

Putting it all together, we have the following behavior of NSF (n).

NSF (n) grows on the order of product of items all on the order of the factorial of n, where the number of elements in the product also grows on the order of a factorial of n.

Cleary, this function grows at an astronomical rate.


I read the letter by I.M.Radu that appeared in [3] stating that there is always a prime between S(n) and S(n+1) for all numbers O<n<4801, where S(n) is the Smarandache function.

Since 1 have a computer program that computes the values of S(n), I decided to investigate the problem further. The serch was

conducted up through n<1,033,197 and for instances where there is

no prime p, where S(n)<ps<S(n+1) . They are as follows: n=224=2:2:2:2'27 S(n) =8 n=2252=3°3°5'5 S(n) =10

nm=2057 =1111°17 S(n) =22 n=2058=2:37'°77 S$(n) =21


27265225 75°5°203'103 S(n) =206 m=265226=2'13°1017101 S(n) =202

N= 843637 =37°151°251 S(n) =302

2=843638 =2°19°149°149 S(n) =298

AS can be seen, the first two values contradict the assertion made by I.M.Radu in his letter. Notice that the last two cases

involve pairs of twin primes. This may provide a clue in the search

for additional solutions.


In [4] T.Yau poses the following problem:

For what triplets n, n+l and n+2 does the Smarandache function

satisfy the Fibonacci relationship S(n)+*S(n+1) = S(n+2) ? Two sciutions S$(9)+S5({(10) = S5(11) 6+5 = 11

S(119) +9(120) = 5(121) 17+5 = 22 were found, but no general solution was given.

To further investigate this problem, a computer program was written that tested all values for n up to 1,000,000. Additional Solutions were found and all known solutions with their prime

factorizations appear in the table below.

S(9)+S(10) =S(11) 9 =3°3 10

2°5 11 = 11

S(119) + §(120) = S(121) 119 = 7-17 120

27235. T21 = 11-11


$(4900) + $(4901) = $(4902); $(26243) + S (26244) = S$(26245)

$(32110) + S$(32111) = S$ (32112) ; S{64008) + $(64009) = S$ (64010) ; $ (368138) + $(368139) = $(368139); S$(415662) + $(415663) = S (415664) ;

I am unable to discern a pattern in these numbers that would lead to a proof that there is an infinite family of solutions.

Perhaps another reader will be able to do so.



The most unsolved problems of the same subject are related to

the Smarandache function in the Analytic Number Theory:

g:Z---N , S(n) is defined as the smallest integer such

that S(n)! is divisible by n.

The number of these unsolved problems concerning the function is equal to... an infinity!! Therefore, they will never be all solved!

One must be very careful in using such arguments when dealing with infinity. As is the case with number theoretic functions, a result in one area can have many aplications to other problems. The most celebrated recent instance is the "prof" of "Fermat's Last Theorem". In this case a result in elliptical functions has the proof as a consequence.

Since S(n) is still largely unexplored, it is quite possible


chat the resolution of one problem leads tc the resolution of many, perhaps infinitely many, others. If that is indeed the case, then

ail problems may eventually be resolved.


1. R. Muller : Unsolved Problems Related to Smarandache Function (Number Theory Publishing Co., Glendale, Az, USA, 1993).

2. P Gronas : A note on S(p‘) (Smarandache Function J., V. 2-3, Nr.1, (1993), 33).

3. M. Radu : Letter to the Editor (Math Spectrum, V.27, Nr.2, (1994/95), 44).

4. T. Yau : A probiem Concerning the Fibonacci Series

(Smarandache Functicn J., V. 4-5, Nr.1, (1994), 42).

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