THE AVERAGE LARGEST PRIME FACTOR

Let P (n) denote the largest prime factor of an integer n. In this paper we look at the average of P (n), and prove that 1 x X nx P (n) = lig(x) + O✏ ⇣ xe c(log x) 3/5 ✏ ⌘ , where lig(x) = R x 2 t x [x/t] log t dt. This improves a result of De Koninck and Iv́ıc and allows us to deduce their asymptotic expansion 1 x X nx P (n) = c0 x log x + c1 1!x (log x) + · · · + ck 1 (k 1)!x (log x) + O x (log x) ! as a corollary, with the advantage that we can give the constants explicitly as cn = 1 2n+1 n X j=0 2j( 1)j⇣(j) (2) j! .

log t dt.This improves a result of De Koninck and Ivíc and allows us to deduce their asymptotic expansion 1 x as a corollary, with the advantage that we can give the constants explicitly as

Introduction
Let P (n) denote the largest prime factor of an integer n.The moments of P (n) were first looked at in 1976 by Knuth and Pardon [4], and concerning the average, they proved that The above asymptotic also appeared in Erdős and Alladi's 1977 paper [1] on additive arithmetic functions.Later, De Koninck and Ivić [2] proved that P nx P (n) has an asymptotic expansion of the form

◆◆
where the constants d i are computable, but not given explicitely.This expansion appears again in [3], where Ivić finds a similar formula for the k th largest prime factor.In this paper, we calculate the average of P (n) up to an error of the form O ✏ ⇣ xe c(log x) 3/5 ✏ ⌘ , like that of the prime number theorem.This allows us to deduce De Koninck and Ivić's expansion as a corollary, as well as give the constants d i explicitely.Our main result is: For any integer k, li g (x) has the asymptotic expansion where These constants c n satisfy Notice in particular that c 0 = ⇡ 2 12 , so we deduce that Although this expansion has previously been presented in [2], and again in [3], there are some advantages to the result above.We are able to give the constants c n explicitely in terms of the zeta function without much additional work, and our proof is basic requiring only an application of the hyperbola method.Furthermore, we give the explicit integral form which has an error term like that of the prime number theorem, paralleling how ⇡(x) is approximated by the function li(x).

The Main Theorem
For each integer n  x, there is at most one prime p > p x such that p|n, and for each prime p there will be exactly h x p i integers less than x which are divisible by p. Combining these two facts, we see that and since By writing h x p i = P nx: p|n 1, and rearranging the order of summation, we obtain Letting K(x) = P px p, we have the estimate which follows from the prime number theorem (See [5]) along with partial summation.Using (7) along with equations ( 5), and (6), we have that which proves (2).To recover the asymptotic expansion in (3), we turn our attention to this integral function and make the substitution t = x u to obtain We can rewrite the integral above as and for any integer k 1, by the geometric series expansion Each of these integrals is absolutely convergent on [1,1), and so the last term contributes an error term of at most and applying the bound To evaluate these constants c n , we use exponential generating series.Integration by parts tells us that for s > 1 and so Multiplying the power series expansions for ⇣(2 z) and 1 2 z yields 0 and hence c n = 1 2 n+1 n X j=0 2 j ( 1) j ⇣ (j) (2) j! .
To find the size of c n we note that ( 1) j ⇣ (j) (2) (j 2 log x) log j 1 x x 3 {x} dx.
Substituting x = e u , the first integral becomes (j + 1), and the second is and so c n = 1 + O n 2 n .This establishes all of Theorem 1.