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I have a reference request related to the result :

$\sum_{n=2}^{x} P(n)$ ~ $\frac{\pi^2}{12}\frac{x^{2}}{log(x)}$ as $x \rightarrow \infty$

where $P(n)$ is the largest prime factor of the positive integer $n$.

Theorem 1.1 in [1] below appears to be the first proof of this result and the result was generalized in Theorem 3.1 in [2] below.

I am looking for upper and lower bounds on the ratio L/R where L and R are the left and right sides respectively of the asymptotic relation above.

Thanks for any help.

References

[1]. K.Alladi and P.Erdos. Pacific J. Math. 71(1977) 275-294

[2]. J.De Konnick and R.Sitaramachandrarao. Indian J. Pure Appl Math. 19(10) 990-1004 Oct 1988

Disclosure: This question was first posted on Math Stack Exchange where it has languished for the last three weeks without an answer (or even a comment).

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  • $\begingroup$ I don't see effective estimates; those are rare. The first edition of Handbook of Number Theory by Mitrinovich, Sandor, and Crstici point to a 1993 paper by J. Lin with a more precise estimate than yours, but still has an error term with unspecified constant. You might check the second edition. $\endgroup$
    – Will Jagy
    Commented Sep 23, 2014 at 2:40
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    $\begingroup$ Actually [1] predates the Alladi--Erdos paper; see: A. E. Brouwer, Two number theoretic sums, Mathematisch Centrum, Amsterdam, 1974, Mathematisch Centrum, Afdeling Zuivere Wiskunde, ZW 1974. $\endgroup$
    – so-called friend Don
    Commented Sep 23, 2014 at 2:42
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    $\begingroup$ It may be worth having a look at Naslund, The Average Largest Prime Factor, Integers 13 (2013) #A81, available online from integers-ejcnt.org/vol13.html $\endgroup$
    – Gerry Myerson
    Commented Sep 23, 2014 at 6:35
  • $\begingroup$ Crossposted, here is the link: math.stackexchange.com/questions/914447/…. $\endgroup$
    – Dietrich Burde
    Commented Sep 23, 2014 at 8:18

1 Answer 1

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I am not exactly sure what you mean by bounds for $L/R$, but I assume that this translates to the rate of convergence and the next terms in the asymptotic. In this short note, a more precise asymptotic for $\sum_{n\leq x} P(n)$ is given and it is shown that $$\frac{1}{x}\sum_{n\leq x}P(n)=\text{li}_g(x) +O_\epsilon \left(x e^{-c(\log x)^{3/5-\epsilon}}\right),$$ where $$\text{li}_g(x)=\int_2^x \frac{t}{x}\frac{[x/t]}{\log t}dt$$ is an integral function that shares some properties in common with $\text{li}(x)=\int_2^x \frac{1}{\log t}dt.$ In particular, we have the asymptotic expansion $$\text{li}_g(x)= \frac{c_0 x}{\log x}+\frac{1!c_1 x}{\log^2x}+\cdots+\frac{(k-1)!c_{k-1} x}{\log^{k}x}+O_k\left(\frac{x}{\log^{k+1}x}\right)$$

where $$c_k=\frac{1}{2^{k+1}}\sum_{j=0}^n \frac{2^j(-1)^j\zeta^{(j)}(2)}{j!},$$ which yields an asymptotic expansion for $\sum_{n\leq x} P(n).$

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  • $\begingroup$ I would interpret the original poster's request as: interpret the LHS as L(x), the RHS as R(x), and ask how far L(x)/R(x) strays from 1. Using your first displayed expression, asymptotically it seems the variance is like e^stuff/(log x)^2, where stuff is the negative of the exponent in your O_epsilon term. It would be interesting to find for which x L(x)/R(x) strays furthest. Gerhard "Likes Taking Things To Extremes" Paseman, 2015.09.03 $\endgroup$
    – Gerhard Paseman
    Commented Sep 4, 2015 at 0:30
  • $\begingroup$ @Gerhard Paseman : Your interpretation of my question is the one I intended. An example of the sort of thing I am seeking is in Rosser and Schoenfeld 1962, Theorem 4, where upper and lower bounds on L/R are given for Chebyshev’s asymptotic relation $\vartheta(x)$ ~ $x$ where $\vartheta(x)=\sum_{p\leq x}\log p$. Those authors show for x ≥ 563, L/R > 1-1/2log(x) and for x > 1, L/R < 1+1/2log(x). I had a look at Eric Naslund’s paper, as suggested by Gerry Myerson in his comment above, but I could not see how to derive bounds on the L/R in my question because of the presence of the big-ohs in the $\endgroup$
    – gjh
    Commented Sep 14, 2015 at 11:05
  • $\begingroup$ @gjh You should edit the question to include these additional remarks. $\endgroup$
    – Neil Strickland
    Commented Sep 14, 2015 at 15:40

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