Is this observation equivalent to Legendre's conjecture? $\forall n \in \Bbb N^+, \exists k \in \Bbb N^+:\ \ (n^2)\# \lt k! \lt (n+1)^2\#$

Question

Is this observation equivalent to Legendre's conjecture?

$$E1:\forall k \in \Bbb N^+ \not \exists n,m \in \Bbb N^+, n \not = m:\ \ \ k! \le n^2\# \lt m^2\# \le (k+1)! $$

This observation is a consequence of some tests regarding the kind answer of @Will Jagy to a previous question at MSE.

$E1$ can be also seen as follows:

$$E2:\forall n \in \Bbb N^+, \exists k \in \Bbb N^+:\ \ (n^2)\# \lt k! \lt (n+1)^2\#$$

Basically, the primorial of a perfect square is always between two factorials $[k,(k+1)!]$, and there not exist another perfect square whose primorial is also inside that same interval. Or in other words: there is at least one factorial between the primorials of two consecutive perfect squares.

I have been able to test up to $100^2$, after that point my computer is quite slow when making the calculation of $(n^2)\#$.

This is the Python code I am using (is not very enhanced, it could be faster, I will try also to make the PARI/GP version). Please use and modify it as you wish:

from sympy import factorial, isprime

lop=[] #list of primes
lof=[] #list of factorials

# create the list of primes
nlimit = 10000
for n in range(1,nlimit):
    if isprime(n):
        lop.append(n)
    lof.append(factorial(n))

for n in range(1,101):
    elemcuad = n * n

    # Calculation of the primorial of n^2
    primoelem = 1
    for k in lop:
        if k<elemcuad:
            primoelem = primoelem * k
        else:
            break

    # Calculation of the closest factorial smaller than or equal to the primorial of n^2
    factant = 1
    for t in range(1,nlimit):
        factt = lof[t-1]
        if factt <= primoelem:
            factant = t
        else:
            break

    # Calculation of the closest factorial greater than the primorial of n^2
    factpost=1
    for t in range(factant,factant+30):
        factt = lof[t-1]
        if factt>primoelem:
            factpost = t
            break

    print("[(n),(n+1)^2] = ["+str((n-1)**2)+","+str(elemcuad)+"] : "+str(factant)+"! <= " + str(elemcuad) + "^2# < " +str(factpost)+"!")

The output starts as follows:

$1! \le 1 \# \lt 2!$

$3! \le 4 \# \lt 4!$

$5! \le 9 \# \lt 6!$

$7! \le 16 \# \lt 8!$

$11! \le 25 \# \lt 12!$

$14! \le 36 \# \lt 15!$

$19! \le 49 \# \lt 20!$

$...$

Its relationship with the first Chebyshev function is this expression:

$$E3: \forall n \in \Bbb N^+, \exists k \in \Bbb N^+, e^{\vartheta (n^2)} \lt k! \lt e^{\vartheta(n+1)^2}$$

or these other ones:

$$E4: \forall n \in \Bbb N^+, \exists k \in \Bbb N^+, \vartheta (n^2) \lt Ln(k!) \lt \vartheta(n+1)^2$$

$$E5: \forall n \in \Bbb N^+, \exists k \in \Bbb N^+, \vartheta (n^2) \lt \sum_{i\ge1} \sum_{j\ge1} Ln((\frac{k}{i}\#)^{\frac{1}{j}}) \lt \vartheta(n+1)^2$$

Basically because $\vartheta(x)=Ln(x\#)$ and $Ln(k!)=\sum_{i\ge1}\psi(\frac{1}{i}k)=\sum_{i\ge1} \sum_{j\ge1} Ln((\frac{k}{i}\#)^{\frac{1}{j}})$.

I would like to ask the following questions:

1. Is it trivial? Are the calculations correct or there is a counterexample? If somebody could kindly test greater intervals would be very appreciated.

2. Are $E1,E2,E3,E4,E5$ equivalent to Legendre's conjecture or are they a stronger conjecture?

Answer 1

If there is a prime between two squares $n^2,(n+1)^2,$ then the primorial $(n+1)^2\#$ will be greater than $n^2\#$ and otherwise not.

So the existence of a factorial (strictly) between primorials of all consecutive squares implies Legendre's conjecture. That's immediate.

So E2 implies Legendre's conjecture.

If Legendre's conjecture is true it is hard to see why that would imply E2 and numbers alone are not very persuasive.