The following $f(m,n)$ function provides the complete set of Euler primes (OEIS A196230):
$$f(m,n)=m^2-m+[\lfloor E^{2^n} \rfloor - {\lfloor E^{2^{n-1}} \rfloor}^2 +\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)}]$$
$$E = 2.893392257682316134127494663$$
$$n\in[0,5] \in \Bbb N\ \ \ \ m = [1..(\lfloor E^{2^n} \rfloor - {\lfloor E^{2^{n-1}} \rfloor}^2+\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)})] \in \Bbb N$$
$E$ is a Mills-like constant I have calculated which is related with Euler's "Lucky" numbers (OEIS A014556, $n$ such that $x^2-x+n$ is prime for $m=0..n-1$). Initially I did not find this constant in any bibliography or reference, and initially it seemed a nice use of a Mills-like constant. So here are the calculations (the questions are at the end):
Def. Euler's "Lucky" constant $E = 2.893392257682316134127494663$
This constant is such that:
$\lfloor E^{2^n} \rfloor -(\lfloor E^{2^{n-1}} \rfloor)^2+\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n}$
for $n=[0..5]$ provides the sequence of Euler's "Lucky" numbers $\{ 2, 3, 5, 11, 17, 41 \}$ in the same growing order.
Euler's lucky constant $E$ is a Mill's like constant obtained by the encapsulation of Euler's "Lucky" numbers into integers $E_n$ of $[N,(N+1)^2]$ intervals. This provides a representing function of Euler's "Lucky" numbers for $n=[0..5]$.
The calculation of the sequence of elements that will be the lower bounds of the $[N,(N+1)^2]$ intervals is as follows:
$E_0=(2)$
$E_1=3+1+({E_1}^2) = 8$
$E_2=5+1+({E_2}^2) = 70$
$E_3=11+1+({E_3}^2) = 4912$
$E_4=17+1+({E_4}^2) = 24127762$
$E_5=41+1+({E_5}^2) = 582148899128686$
$E=E_5^{\frac{1}{2^5}}=2.893392257682316134127494663$
So we can obtain Euler's "Lucky" numbers as follows:
$Lucky_{n+1}=\lfloor E^{2^n}\rfloor-\lfloor E^{2^{n-1}}\rfloor^2+ \frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)}$
$Lucky_1=\lfloor E^{2^0}\rfloor-\lfloor E^{2^{0-1}}\rfloor^2+ \frac{\lvert 0-(\frac{1}{2}) \rvert}{(\frac{1}{2})-0)}=2$
$Lucky_2=\lfloor E^{2^1}\rfloor-\lfloor E^{2^{1-1}}\rfloor^2+ \frac{\lvert 1-(\frac{1}{2}) \rvert}{(\frac{1}{2})-1)}=3$
$Lucky_3=\lfloor E^{2^2}\rfloor-\lfloor E^{2^{2-1}}\rfloor^2+ \frac{\lvert 2-(\frac{1}{2}) \rvert}{(\frac{1}{2})-2)}=5$
$Lucky_4=\lfloor E^{2^3}\rfloor-\lfloor E^{2^{3-1}}\rfloor^2+ \frac{\lvert 3-(\frac{1}{2}) \rvert}{(\frac{1}{2})-3)}=11$
$Lucky_5=\lfloor E^{2^4}\rfloor-\lfloor E^{2^{4-1}}\rfloor^2+ \frac{\lvert 4-(\frac{1}{2}) \rvert}{(\frac{1}{2})-4)}=17$
$Lucky_6=\lfloor E^{2^5}\rfloor-\lfloor E^{2^{5-1}}\rfloor^2+ \frac{\lvert 5-(\frac{1}{2}) \rvert}{(\frac{1}{2})-5)}=41$
The term
$$\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)}$$
is just a correction (gently provided by @Steven Harding in this question) to calculate properly the initial case $Lucky_1$ associated with $n=0$ and join it to the rest of cases. The value $E_0$ is an special case because there is not previous term.
$\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)}=+1$ for $n=0$ and $=-1$ for the rest of cases.
Finally, replacing the expression into the Euler's "Lucky" polynomial, the result is the two variables function able to reproduce the complete set Euler primes (OEIS A196230):
$$f(m,n)=m^2-m+[\lfloor E^{2^n} \rfloor - {\lfloor E^{2^{n-1}} \rfloor}^2 +\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)}]$$
$$E = 2.893392257682316134127494663$$
$$n\in[0,5] \in \Bbb N\ \ \ \ m = [1..(\lfloor E^{2^n} \rfloor - {\lfloor E^{2^{n-1}} \rfloor}^2+\frac{\lvert n-(\frac{1}{2}) \rvert}{(\frac{1}{2})-n)})] \in \Bbb N$$
As a matter of curiosity proposed the constant decimal expansion as a new sequence at OEIS, and professor Sloane wisely considered that the use of a generating function has sense only for infinite sequences, so it does not make sense using a Mills-like constant to create a representing function for so very few terms because the expression is longer than the terms to be produced (the "Lucky" numbers $\{ 2, 3, 5, 11, 17, 41 \}$). I understood the reasons, but having that in mind, the generation of a two variable function able to produce the whole set of $57$ "Lucky" primes seemed interesting to me.
I would like to ask the following question:
Are the calculations correct? Did I forget something?
Does it make sense an expression like this using a mathematical tool like Mills' constant for a finite sequence or is just a curious artificial "Frankenstein"-expression? Initially I understand that it is a curiosity only, but looking at the function it is interesting to see that there are two second degree terms. One depends on the variable $m$, the other one depends on the value $n$ associated to the constant $E$, so some diophantine relationship is there that could be explored. Thank you!