Erdős Problem 648: decreasing greatest prime factors
Statement and claim.
Write $P(a)$ for the greatest prime factor of $a\geq2$. Let $g(n)$ be the largest length of a sequence \[
2\leq a_1<\cdots<a_t<n,\qquad P(a_1)>\cdots>P(a_t).
\] We reconstruct Cambie's determination of the order of magnitude: \[
(2-o(1))\sqrt{\frac n{\log n}}
\ \leq g(n)\ \leq\
(2\sqrt2+o(1))\sqrt{\frac n{\log n}}.
\] The source is Stijn Cambie,
On Erdős problem #648,
https://arxiv.org/abs/2503.22691; see also
https://www.erdosproblems.com/648. The only asymptotic input below is the prime number theorem. The existence or value of a limiting leading constant is not asserted.
Upper bound.
Put $p_i=P(a_i)$ and $b_i=a_i/p_i$. Since $a_{i+1}>a_i$ and $p_{i+1}<p_i$, we have $b_{i+1}>b_i$. These are positive integers, so $b_i\geq i$. If $q_j$ is the $j$-th prime, then the $t-i+1$ distinct primes $p_i,\ldots,p_t$ show that $p_i\geq q_{t-i+1}$. Therefore \[
n>a_i=b_ip_i\geq i q_{t-i+1}.
\] Taking $i=\lfloor t/2\rfloor$, and using $q_j\sim j\log j$, gives \[
n\geq(1-o(1))\frac{t^2\log t}{4}.
\] The lower bound proved next shows that the extremal $t=g(n)$ is at least a constant times $\sqrt{n/\log n}$. The displayed upper bound then implies $\log t=(\tfrac12+o(1))\log n$, and hence the claimed upper constant $2\sqrt2$.
Lower bound.
Fix $0<\varepsilon<1$, and put $z=(1-\varepsilon)\sqrt{n\log n}$. List all primes in $(\sqrt n,z]$ in decreasing order as $p_1>\cdots>p_r$. Starting with $a_0=0$, define \[
a_i=p_i\left(\left\lfloor\frac{a_{i-1}}{p_i}\right\rfloor+1\right).
\] Then $a_i>a_{i-1}$ and \[
a_i\leq\sum_{j=1}^i p_j\leq\sum_{p\leq z}p
\sim\frac{z^2}{2\log z}
\sim(1-\varepsilon)^2n<n.
\] Here the prime-sum estimate follows from the prime number theorem by partial summation. Since $p_i>\sqrt n$ and $a_i<n$, the cofactor $a_i/p_i$ is smaller than $\sqrt n<p_i$. All its prime factors are therefore smaller than $p_i$, proving $P(a_i)=p_i$. Finally, \[
r=\pi(z)-\pi(\sqrt n)
=(2(1-\varepsilon)+o(1))\sqrt{\frac n{\log n}}.
\] Letting $\varepsilon$ tend to zero gives the lower bound.
Scope.
This proves the known order-of-magnitude resolution, including the two displayed asymptotic constants, rather than an exact formula for $g(n)$. It is an attributed reconstruction, not a new resolution.
Completion estimate.
COMPLETION: 100%