# Proof that the error between $W(\lambda(\sqrt{n}))$ and the fraction of composite integers in $\mathbb{O}_n$ declines as $n$ increases.

Please provide assistance for the proof that the error between the function $$W(\lambda(\sqrt{n}))$$ and the fraction of composite integers in $$\mathbb{O}_n$$ declines as $$n$$ increases.

Let $$\mathbb{O}_n$$ represent the set of odd integers less than or equal to $$n$$ excluding 1.

$$\mathbb{O}_n$$ = {3,5,7,9,11,13,15,17,19,…n}

Let the function $$l(x)$$ equal the largest prime number less than $$x$$.

For example: $$l(10) = 7, l(20) = 19$$ and $$l(23) = 19$$

Let the function $$\lambda(x)$$ equal the largest prime number less than or equal to $$x$$.

For example: $$\lambda(10) = 7, \lambda(20) = 19$$ and $$\lambda(23) = 23$$

The function $$W(\lambda(\sqrt{n}))$$ was derived to represent the fraction of composite integers in the set $$\mathbb{O}_n$$ for large values of $$n$$.

$$\begin{eqnarray*} W(\lambda(\sqrt{n})) = \sum_{\substack{p=3\\p\ \text{prime}}}^{\lambda(\sqrt{n})}\left((\frac{1}{p})\prod_{\substack{q=3\\q\ \text{prime}}}^{l(p)}\frac{(q-1)}{q}\right) \end{eqnarray*}$$ or

$$W(\lambda(\sqrt{n}))= \frac{1}{3} + (\frac{1}{5})(\frac{2}{3}) + (\frac{1}{7})(\frac{2}{3})(\frac{4}{5}) + ... + (\frac{1}{ \lambda(\sqrt{n})})(\frac{2}{3})(\frac{4}{5})(\frac{6}{7})(\frac{10}{11})...(\frac{(l(\lambda(\sqrt{n})) - 1)}{l(\lambda(\sqrt{n}))})$$.\

I have proven that the maximum error between $$W(\lambda(\sqrt{n}))$$ and the actual fraction of composite integers in the set $$\mathbb{O}_n$$ declines as $$n$$ increases in sections 6 and 7 in this paper http://vixra.org/pdf/1901.0436v5.pdf.

Is this proof sufficient? If not, what needs to be done?

Let me clean it up. Let $$lpf(n)$$ the least prime factor and $$\mu(d)$$ the Möbius function and as usual $$p,q$$ primes then $$\sum_{n \le x, lpf(n)=p} 1 = \sum_{d | \prod_{q < p} q} \mu(d) \lfloor \frac{x}{pd}\rfloor$$

What you did is approximating $$\lfloor \frac{x}{pd}\rfloor \approx \frac{x}{pd}$$ obtaining $$\sum_{n \le x, lpf(n)=p} 1 \approx \sum_{d | \prod_{q < p} q} \mu(d) \frac{x}{pd}= \frac{x}{p}\sum_{d | \prod_{q < p} q} \frac{\mu(d)}{d}= \frac{x}{p}\prod_{q < p} (1-\frac1q)$$

Then we have $$\lfloor x \rfloor = \sum_{p \le x}\sum_{n \le x, lpf(n)=p} 1$$ $$\lfloor x \rfloor-\pi(x) = \sum_{p \le \sqrt{x}}(-1+\sum_{n \le x, lpf(n)=p} 1)$$

and $$1-\sum_{p \le x} \frac{1}{p}\prod_{q < p} (1-\frac1q) = \prod_{q \le x} (1-\frac1q)$$

$$\sum_p \frac{1}{p}\prod_{q < p} (1-\frac1q)=1$$

To go further you need to estimate those things using the Mertens theorems, the PNT, the RH.

I identified a flaw in determining the maximum error between the $$W()$$ function and the fraction of composite numbers less than $$n$$.

For example, the fraction of odd composite numbers less than $$n$$ that are divisible by 7 (excluding 7) and not divisible by a lower prime is estimated as (1/7)(2/3)(4/5).

The maximum error due to this estimation was precisely determined for each of the fractions (e1, e2, e3).

1/7 ± e1

2/3 ± e2

4/5 ± e3

These errors were incorrectly combined as:

$$e_{max}^7 = (1/7)(2/3)(4/5)(e1/(1/7) + e2/(2/3) + e3/(4/5))$$

The correct maximum possible error is for composite numbers divisible by 7 is:

$$e_{max}^7 = (1/7 + e1)(2/3+e2)(4/5+e3) - (1/7)(2/3)(4/5)$$

The difference in the results due to this error is negligible, but for a mathematical proof, any error is not acceptable. The error was corrected, and a new revision of the paper is now available.

http://vixra.org/pdf/1901.0436v6.pdf