Can we replace the upper limit condition of the Sieve of Eratosthenes $\sqrt{n}$ with the value $\sqrt{p}$ where $p$ is the last sieved prime $\lt n$?

Question

By chance I stumbled upon the OEIS list A033677 of the smallest divisor of $n$ greater or equal to $\sqrt{n}$. Roughly speaking if we use the classic enhanced sieve of Eratosthenes, $\sqrt{n}$ is the typical upper limit to find a divisor $d \gt 1$ of $n$, so if $n$ is composite then it is expected that $\exists d \in [2,\sqrt{n}]$.

I did the same exercise but instead of $\sqrt{n}$ using as the limit the square root of the last sieved prime, which is indeed the previous closest prime $p$ to $n$ and made the calculation of the smallest divisor of $n$ greater or equal to $\lceil \sqrt{p} \rceil$.

Testing with Python and if I did not make an error, curiously I obtained exactly the same list than A033677 for every $n \in [3,6\cdot 10^6]$ (after that point it gets slower, I will try with PARI\GP later). So I wonder if it is possible to use the above mentioned $\lceil \sqrt{p} \rceil$ upper limit (the ceil of the square root of the last sieved prime) instead of $\sqrt{n}$ to find a divisor of $n$, if any.

This is the (very basic) Python code that compares one by one the elements of the sequence A033677 with the sequence of smallest divisor of $n$ greater or equal to $\sqrt{p}$, please feel free to use it:

from sympy import prevprime, divisors
    for n in range (3,10000000):
        A0033677_Value=0
        for d in divisors(n):
            if d>=sqrt(n):
                A0033677_Value=d
                break

        for d in divisors(n):
            if d>int(sqrt(prevprime(n))):
                if A0033677_Value == d:
                    print("SUCCESS "+str(n))
                else:
                    print("ERROR " + str(n) + "\t" + str(A0033677_Value)+"\t"+str(prevprime(n))+"\t"+str(d))
                    return
                break

If the observation is true, the property would be something like this:

$\forall n$ (composite)$\ \exists d: d|n\ , d \in [2,\lceil \sqrt{p_1} \rceil], p_1 \in \Bbb P, p_1\lt n,\ \land \not\exists\ p_2: p_1 \lt p_2 \lt n$.

For instance for $n=20,21,22,23$ we would look for a divisor of $n$ in the interval $[2,\lceil \sqrt{19} \rceil]$ because $19$ is the last sieved prime before $n$. As we already sieved $19$ in a former step it is in our list of already available primes and thus we can used (meaning we already know at this point that $19$ is prime). Then applying Eratosthenes as usual it would be detected that $n=20,21,22$ are composite and $n=23$ is the next prime.

UPDATE: as kindly explained by @Slade in the comments, in some cases (e.g when $n$ is a perfect square) having the same divisor $d$ over $\sqrt{n}$ and $\sqrt{p}$ does not imply that we can find a divisor at the interval $[2,\sqrt{p}]$ and it is required to go up to $[2,\lceil \sqrt{p}\rceil]$, meaning that the first divisor is exactly $\lceil \sqrt{p}\rceil$. That is the reason why the ceil function is used to set the upper limit.

If true I am not sure if it might be significant. By Bertrand's postulate, there should be a prime between $[\frac{n}{2},n]$, so in the best of cases, the previous prime would be $p_i=\frac{n}{2}$, so it would imply for that best of cases the use of $\sqrt{\frac{n}{2}}$ as the upper limit for the sieve instead of $\sqrt{n}$.

I would like to ask the following questions:

1. Would it be possible to use $\lceil \sqrt{p} \rceil $ instead of $\sqrt{n}$? Was already in use that approach?

2. Is there a counterexample (probably very big)? If not, what could be the reason behind it? Thank you!

UPDATE: the code below tests the Sieve of Eratosthenes using the modified algorithm as explained above, including the special condition for perfect squares. It seems to work so far up to $10^5$. I will test further.

from gmpy2 import is_prime, is_square
    # modified Eratosthenes, upper sieving limit sqrt(p)
        lopM=[2] #list of already sieved primes
        lp=2 # last prime sieved
        for pos in range(3,1000008):
            ul=int(sqrt(lp)) #upper limit
            if is_square(pos):
                ul=ul+1 # correction for perfect squares
                # basically: if is a perfect square we would continue for
            if ul==1:
                ul=2 # exclude 1 of possible divisors
            comp=False #composite detected
            for j in lopM: # using the list of already known primes
                if j in range(2,ul+1): # while j is under the upper limit
                    if (pos%j)==0:
                        comp=True
                        break
                else: # we arrived to the upper limit
                    break
                if comp==False:
                    lopM.append(pos)
                    lp=pos

        for elem in lopM:
            if not is_prime(elem): # sanity check
                print("ERROR, not prime " + str(elem))
                return
        print("Primes sieved: " + str(len(lopM)) + " and last prime was " + str(lp))

Answer 1

1) As other people stated, imagine $n=q^2$, $q$-prime, then $p_{\pi(q^2)} < q^2 < p_{\pi(q^2)+1}$ which is $\sqrt{p_{\pi(q^2)}} < q < \sqrt{p_{\pi(q^2)+1}}$. Bertrand's postulate is not going to help, because if you stick to any bound $\leq \sqrt{p_{\pi(q^2)}}$, the algorithm will tell you $q^2$ is prime.

2) If $n=q\cdot b$, $q$-prime and $q<b$, then $p_{\pi(n)} < q\cdot b < p_{\pi(n)+1}$. Assuming Oppermann's conjecture is true (some references here), then $q<\sqrt{p_{\pi(n)}}$, otherwise (suppose contrary) if $$\sqrt{p_{\pi(n)}} \leq q \Rightarrow p_{\pi(n)} \leq q^2 < q \cdot (q+1) \leq q\cdot b < p_{\pi(n)+1}$$ and according to that conjecture there will be a prime between $q^2$ and $q \cdot (q+1)$. This contradicts the fact that $p_{\pi(n)+1}$ is the next prime after $p_{\pi(n)}$. So, technically, you can lower the bound in this particular case and assuming Oppermann's conjecture (!)