The asymptotic order of the logarithm of the square-free part of n! is shown to be (log 2)n with error O(√n )

Broughan, Kevin A.

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Research Commons@Waikato

ASYMPTOTIC ORDER OF THE SQUARE-FREE PART OF N!Kevin A. BroughanDepartment of Mathematics, University of Waikato, Hamilton, New Zealandkab@waikato.ac.nzReceived: 4/27/02, Revised: 7/26/02, Accepted: 7/26/02, Published: 7/29/02AbstractThe asymptotic order of the logarithm of the square-free part of n! is shown to be (log 2)nwith error O(pn).1. IntroductionIf the standard prime factorization of n! is considered over a range of values of n then anumber of patterns are apparent:10! = 28 ¢ 34 ¢ 52 ¢ 720! = 218 ¢ 38 ¢ 54 ¢ 72 ¢ 11 ¢ 13 ¢ 17 ¢ 1930! = 226 ¢ 314 ¢ 57 ¢ 74 ¢ 112 ¢ 132 ¢ 17 ¢ 19 ¢ 23 ¢ 2940! = 238 ¢ 318 ¢ 59 ¢ 75 ¢ 113 ¢ 133 ¢ 172 ¢ 192 ¢ 23 ¢ 29 ¢ 31 ¢ 37:All the primes up to n appear. If p and q are primes appearing in the factorization withp < q and ®; ¯ are the highest powers of p and q dividing n! respectively, then ® ¸ ¯, i.e.the smaller the prime, the larger the power. Even though sometimes a given power doesnot appear (the power 3 is missing from 20! even though the powers 2 and 4 appear),the power 1 always appears.The square-free part of n! is the number a, with no square factors, which appears inthe factorizationn! = ab2:It is easy to see that a is exactly the product of each of the primes which appear toan odd power in the standard factorization, and in particular is divisible by the primesappearing to power 1 in that factorization.Two natural questions arise: what is the size of the square-free part a of n! and whatproportion of a is the product of the primes which occur to power 1 ? In this note itINTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002),#A10 2will be shown that, asymptotically, the square-free part of n! has order 2n and that theproportion of primes to power 1 is about 72%.2. Integer Square RootsFor each whole number n let the integer lower square root be de¯ned byr¡(n) =Yp®jjnpb®2cand the integer upper square root byr+(n) =Yp®jjnpd®2e:If n = ab2 and cn = d2 with a and c square-free, thenb = r¡(n); d = r+(n); a = c =r+(n)r¡(n):This pair of functions r§ is quite useful. They are multiplicative, can be generalizedto integer k'th roots and are related to the integer conductor or square-free core. Forexamples and applications see [3, 4].3. Computing the square-free part of n!To obtain some idea of the behavior of the square-free part of n!, for large n, it paysto do some computations. However, for numbers of quite small size, say n = 400, n! isa number with over 800 digits, so ¯nding the square-free part should not be attempteddirectly. The following strategy was adopted:For each n ¸ 1, let µn be the square-free part of n+ 1, i.e.,µn = r+(n+ 1)=r¡(n+ 1):Because an+1b2n+1 = (n+ 1)n! = (n+ 1)anb2n and n+ 1 = µnc2 for some integer c, wehave µnanb2n = an+1b2n+1.If a prime p j (µn; an), then p occurs as a factor in both µn and an, so must occur toan odd power in both n! and n+ 1, and therefore to an even power in (n+ 1)!. Hence itdoes not occur in an+1. If a prime occurs in just one of µn and an, then it must occur inan+1. This leads directly to the formula:an+1 =anµn(an; µn)2:(1)INTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002),#A10 3Note that this formula can be used to evaluate the sequence (an) recursively, so thevalues of log an can be plotted, revealing a nice approximately linear dependence on n.See Figure 1.Figure 1. The sequence log an as a function of n.4. Asymptotic ordersThe result of these computations of the square-free part of n! leads to two natural tasks:determining the slope of a line approximating the graph of log an, and ¯nding an upperbound for the error in this approximation. The completion of both tasks is summarizedin the next theorem.Theorem 1: For each n 2 N let n! = anb2n where a1 = b1 = 1 and where for all n ¸ 1,an is square-free.log an = n log 2 +O(pn);Thenlog bn =12n log n¡ 1 + log 22n+O(pn):andProof: Consider the central binomial coe±cient¡2nn¢= tns2n where tn is square-free.Thenb22na2n = (2n)! = (n!)2s2ntnso tn = a2n for all n 2 N . By the main result in [7], there is a real strictly positiveconstant c such that for all ² > 0 and all n su±ciently large(c¡ ²)pn < 2 log sn < (c+ ²)pn:Therefore log sn = O(pn).INTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002),#A10 4Stirling's approximation for n! [8] is n! ¼ p2¼n(n=e)n. It leads to the formula:log n! = n log n¡ n+O(log n):log a2n = logµ2nn¶¡ 2 log snConsequently:= 2n log 2n¡ 2n¡ 2n log n+ 2n+O(pn)= 2n log 2 +O(pn):log a2n+1 = log a2n + log µ2n ¡ 2 log(a2n; µ2n)By equation (1)= log a2n +O(log n) since µ = O(n)= (2n+ 1) log 2 +O(pn)log an = n log 2 +O(pn):and thereforeBut, by Stirling's approximation again and this estimate for log an:2 log bn = n log n¡ n¡ n log 2 +O(pn)= n log n¡ (1 + log 2)n+O(pn)and therefore log bn =12n log n ¡ 1+log 22n + O(pn). This completes the proof of thetheorem.It follows also that the square-free part of¡2nn¢, namely tn, satis¯es log tn = 2n log 2 +O(pn), giving the asymptotic order. This relates to the solved conjecture of Erd}os [5]that the binomial coe±cient¡2nn¢is not square-free for n > 4. It relates also to the parityof the exponents of the prime factors of n!, [2].5. Primes dividing n!Lemma 1: Let k ¸ 1 and let p be a prime integer. If n ¸ k(k + 1) then pkkn! if andonly if nk+1< p · nk.Proof: If nk+1< p · nkthen k · np< k + 1, so thereforek = bnpc:Since k(k + 1) · n we have k · nk+1< p, so thatb np2c < k + 1p· 1:It follows that b np2c = 0, by Legendre's formula®p =1Xj=1b npjc = bnpc = k:INTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002),#A10 5Conversely, if pkkn! then k = bnpc+ ¢ ¢ ¢ . Thus bnpc · k, which implies nk+1< p, so k < p.In addition k < nk+1, therefore np2· kp< 1 so b np2c = 0 and k = bnpc, which shows p · nk.This completes the proof of the lemma.For x > 0 letµ(x) =X2·p·xlog p;Chebyshev's function [1], where the sum is over all primes less than or equal to x. Ifx ¸ 563 then µ(x) is close to x in that [6]x¡1¡ 12 log x¢< µ(x) < x¡1 +12 log x¢:If follows that if n ¸ nkjµ¡nk¢¡ µ¡ nk + 1¢¡ nk(k + 1)j · nk log nk:By Lemma 1, the logarithm of the product of primes which appear in n! to the k'thpower islogYnk+1<p·nkp =Xnk+1<p·nklog p= µ¡nk¢¡ µ¡ nk + 1¢=nk(k + 1)+Ok¡ nlog n¢;so the asymptotic order of the product is nk(k+1)as n!1.Therefore, by Theorem 1, the asymptotic proportion of the square-free part of n! dueto primes appearing to powers 1; 3; : : : ; 2k ¡ 1 is1¡ 12+ 13¡ 14+ ¢ ¢ ¢+ 12k¡1 ¡ 12klog 2:For example, primes to power one contribute 1=2log 2or about 72%, and those to powerone or three to 7=12log 2, or about 84% of the square-free part.AcknowledgmentsThis work was done in part while the author was on study leave at Columbia University.The support of the Department of Mathematics at Columbia University and the valuablediscussions held with Patrick Gallagher are warmly acknowledged. The contributions ofa referee who, in particular, supplied the general case for Lemma 1, are also gratefullyacknowledged.INTEGERS:ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 2 (2002),#A10 6References[1] T. M. Apostol, Introduction to Analytic Number Theory, New York, Berlin Heidel-berg: Springer-Verlag, 1976.[2] D. Berend, On the parity of the exponents in the factorization of n!, J. NumberTheory 64 (1997), 13-19.[3] K. A. Broughan, Restricted divisor sums, Acta Arithmetica, 101 (2002), 105-114.[4] K. A. Broughan, Relationships between the integer conductor and k'th root functions,(preprint).[5] A. Granville and O. Ramar¶e, Explicit bounds on exponential sums and the scarcityof square-free binomial coe±cients, Mathematica 43 (1996), 73-107.[6] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of primenumbers, Illinois J. Math. 6 (1962), 64-94.[7] A. S¶arkÄozy, On divisors of binomial coe±cients I, J. Number Theory 20 (1985),70-80.[8] http://mathworld.wolfram.com/StirlingsSeries.html.

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arkÄ ozy, On divisors of binomial coe±cients

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Introduction to Analytic Number Theory,

On the parity of the exponents in the factorization of n!,

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Asymptotic order of the square free part of n!

Name not available

The asymptotic order of the logarithm of the square-free part of n! is shown to be (log 2)n with error O( √n)

NEUROSURGERY ENTHUSIASTIC WOMEN SOCIETY

Asymptotic Order of the Square-free Part of N!

Asymptotic order of the square free part of n!

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