4 final ? major round

$P' - T' = 1/(\beta_n z) - e^{\alpha_n}z(n)(1/2)^{2e(\alpha_n e^{\alpha_n}z(1/2)^{2e(\alpha_n + \log(z))}$.

and this gives a value for $l$ which in turn gives $L > 0$. Since for the improvement I assume $n > 30,$ the fudge factor is at most 18/11, and when $\beta_n=3$ and $\alpha_n = 0.75$, the denominator of the last fraction goes from some value above 1/5 to 1/3 as $n$ increases. So the whole expression is bounded by $9 \binom{n+1}{s} z$, or $9 \binom{n+1}{s} \log{p_n}$. Since $\log(\log(p_n))$ is increasing, the whole ball of wax is bounded by $n^{2e(\log\log(p_n) + \alpha_n) +1}\log(p_n}$.1}\log(p_n)$. 3 second major round$\text{ceil}{l/t} \text{ceil}(l/t) \ge J_t \ge \text{floor}{l/t}$. text{floor}(l/t)$. So the count of numbers in $I$ not

Improvement? (Paseman): For sufficiently large $n$, $j(m) \le O^*(n^{(2 +2e \log(\log(p_n))})$.log(\log(p_n)))})$.$\mu(t)$is the M$\"{o}$bius Moebius function, so$\mu(t)$As in the question above, this gives$P = \prod_{1 \le i \le n}(1 - 1/m_i}$1/m_i)$ (Recall that the distinct prime factors of $m$ are the $m_i$.) Then $T = P - T_s$, so

then $T' = e^h(n)2^{-s-1e^{h(n)}2^{-s-1} \gt e^{h(n)}(h(n)^{s+1})/((s+1)!) \gt T$.

Also $P = \prod_{1 \le i \le n}(1 - 1/m_i} ) \gt \prod_{1 \le i \le n}(1 - 1/p_i} ) \gt 1/(\beta_n\log(p_n))$,

$P' -T' = (\beta_n z)^{-1} - (e/2^{2e})^{\alpha_n} z^{-2e\log(2) + 1}$

$= (\beta_n z)^{-1} - ((e^{\alpha_n})z)^{-2e\log(2) + 1}$

(z/[(\beta_n)^{-1} - (e^\alpha_n)^{-2e\log(2) + 1} z^{-2e\log(2) + 2}]) \gt \frac{SB}{P-T}$an estimate on$\prod_{1 \le i \le n} (1 - 1/p_i)$, which is$e^{-(\gamma + \delta)}/\log(p_n)$, where$e^-\gamma$e^{-\gamma}$ is close to 1/1.78 and $\delta$ is bounded by a sum of three terms, the largest of which is again $4/\log(p_n+1)$, again requiring that $p_n + 1$ be bigger than (something close to) $e^8$. However, computationsseem to show that the constants chosen seem to allow the required estimates to hold for $n \gt 30$ and some $n$ smaller than 30. The major block is on $n \gt 2s \gt 2eh(n) - 1$.

2 first major round
then led me to improve Westzynthius's result to $Q*2^g(n)$ Q*2^{g(n)}$where$g(n)$was$n/2 + O(log(n))$O(\log(n))$.

one $a+i \in I$. Let $\text{rad}(m) = \prod_{p \text{prime}, text{ prime,} p \mid m} p$ be the largest$\ceil{l/t} \text{ceil}{l/t} \ge J_t \ge \floor{l/t}$. text{floor}{l/t}$. So the count of numbers in$I$not Theorem (Stevens)$j(m) < 2n^{(2 + 2e\log(n)}$2e\log(n))}$.

Improvement? (Paseman): For sufficiently large $n$, $j(m) = \le O^*(n^{(2 +2e \log(\log(p_n))})$.

the improvement together. In discussing the improvement, I will ask that $n \gt 30$.

$\mu(t)$ is the M\"{o}bius M$\"{o}$bius function, so $\mu(t)$

Where , where

Let's do $SB$. Stevens's replacement is $n^s$; mine is $\binom{n+1)}{s}$ \binom{n+1}{s}$times a small fudge factor,$\binom{n+1}{s-2} / \binom{n+1}{s}$. This gives$(1 1 + \frac{s(s-1)}{(n+2)(n+3 -2s)}$.(s(s-1))/((n+2)(n+3-2s))$.

As above, this gives $P = \prod{1 prod_{1 \le i \le n}(1 - 1/m_i}$ (Recall that the distinct prime

$i! \sum_{t (i!)\sum_{t \mid m , \nu(t) = i} 1/t \le (\sum_{1 \le j \le n} 1/m_j)^i \le h(n)^i$,so $T \lt \sum_{s \lt i \le n} \frac{(\sum_{1 ((\sum_{1 \le j \le n} 1/m_j)^i}{i!}$1/m_j)^i)/(i!)$, so$T \lt \sum_{s \lt i } \frac{h(n)^i}{i!} (h(n)^i)/(i!) \le e^{h(n)}\frac{h(n)^{s+1}}{(s+1)!}$e^{h(n)}(h(n)^{s+1})/((s+1)!)$.

then $T' = e^h(n)2^(-s-1) e^h(n)2^{-s-1} \gt e^{h(n)}\frac{h(n)^{s+1}}{(s+1)!} e^{h(n)}(h(n)^{s+1})/((s+1)!) \gt T$.

Also $P = \prod{1 prod_{1 \le i \le n}(1 - 1/m_i} \gt \prod{1 prod_{1 \le i \le n}(1 - 1/p_i} \gt 1/\beta_n\log(p_n)$1/(\beta_n\log(p_n))$,so let$P' = 1/\beta_n\log(p_n)$1/(\beta_n\log(p_n))$. Again the valid range of $n$ depends on the

$P' - T' = 1/(\beta_n z) - e^{\alpha_n}z(n)(1/2)^(2e(\alpha_n e^{\alpha_n}z(n)(1/2)^{2e(\alpha_n + \log(z)))$.log(z))}$. Now$(1/2)^2e\log(z) (1/2)^{2e\log(z)} = (z)^{-2e\log(2)}$. z^{-2e\log(2)}$. So

$\binom{n+1}{s} ( 1 +\frac{s(s-1)}{(n+2)(n+3 (s(s-1))/((n+2)(n+3 -2s)}) 2s)) )\frac{z}{(\beta_n)^{-1} (z/[(\beta_n)^{-1} - (e^\alpha_n)^{-2e\log(2) + 1} z^{-2e\log(2) + 2}} 2}]) \gt /\frac{SB}{P-T}$and this gives a value for $l$ which in turn gives $L > 0$. Since for the improvement I assume $n > 30, 30,$ the fudge factor is at most 18/11, and when \beta_n=3 $\beta_n=3$ and \alpha_n $\alpha_n = 0.750.75$, the denominator of the last fraction goes from some value above 1/5 to 1/3 as $n$ increases. So the whole expression is bounded by $9 \binom{n+1}{s} z$, or $9 \binom{n+1}{s} \log{p_n}$. Since \log(\log(p_n))$\log(\log(p_n))$ is increasing, the whole ball of wax is bounded by $n^{2e(\log\log(p_n) + \alpha_n) +1}\log(p_n}$.

Now as to the choice of \alpha_n $\alpha_n$ and \beta_n. $\beta_n$. They were chosen generously: Mertens showed that\sum_{1 $\sum_{1 \le i \le n} 1/p_i \lt \log\log(p_n) + B + delta$, \delta$, where$B$is a constant close to 0.26and delta$\delta$is an error bounded in size by a sum of two terms, the largest of which is 4/log(p_n$4/\log(p_n +1).So using Merten's estimate, $p_n + 1 1$ should be bigger than (some number not much larger thane^8. ) $e^8$. Similarly, Mertens hasan estimate on \prod_{1 $\prod_{1 \le i \le n} (1 - 1/p_i)1/p_i)$, which is $e^{-(\gamma + delta)}/\log(p_n), \delta)}/\log(p_n)$, where e^-\gamma $e^-\gamma$ is close to 1/1.78 and delta $\delta$ is bounded by a sum of three terms, the largest of which is again 4/log(p_n+1), $4/\log(p_n+1)$, again requiring that $p_n + 1 1$ be bigger than (something close to) e^8. $e^8$. However, computationsestimates to hold for $n >30 \gt 30$ and some n $n$ smaller than THe The major block is on $n > \gt 2s > \gt 2eh(n) - 11$.

1