Euler's "Lucky" constant

Based on my results in the previous post regarding a prime-representing function based on Bertrand's postulate I was able to produce a new constant related with Euler's "Lucky" numbers. Initially I did not find this constant in any bibliography or reference, and it seems a nice use of a Mills-like constant. So here are the results:

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" prime numbers in 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" primes 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+(E1^2) = 8

E(2)=5+1+(E2^2) = 70

E(3)=11+1+(E3^2) = 4912

E(4)=17+1+(E4^2) = 24127762

E(5)=41+1+(E5^2) = 582148899128686

And finally Euler's "Lucky" constant (E) has the following value:

E=E(5)^(1/(2^5))=2.893392257682316134127494663

The above manipulation encapsulates the "Lucky" primes into a different upper interval, so the following calculation returns the "Lucky" primes back from the representing function:

Lucky(n+1)=floor(E^(2^n))-(floor(E^(2^(n-1))))^2+(abs(n-(1/2))/((1/2)-n))

for n=0..5

Thus:

Lucky(1)=floor(E^(2^0))-(floor(E^(2^(0-1))))^2+(abs(0-(1/2))/((1/2)-0))=2

Lucky(2)=floor(E^(2^1))-(floor(E^(2^(1-1))))^2+(abs(1-(1/2))/((1/2)-1))=3

Lucky(3)=floor(E^(2^2))-(floor(E^(2^(2-1))))^2+(abs(2-(1/2))/((1/2)-2))=5

Lucky(4)=floor(E^(2^3))-(floor(E^(2^(3-1))))^2+(abs(3-(1/2))/((1/2)-3))=11

Lucky(5)=floor(E^(2^4))-(floor(E^(2^(4-1))))^2+(abs(4-(1/2))/((1/2)-4))=17

Lucky(6)=floor(E^(2^5))-(floor(E^(2^(5-1))))^2+(abs(5-(1/2))/((1/2)-5))=41

Q.E.D.

The term (abs(n-(1/2))/((1/2)-n)) is just a correction 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. (abs(n-(1/2))/((1/2)-n))=+1 for n=0 and =-1 for the rest of cases.

Finally, the following polynomial in two variables provides the complete set Euler primes (A196230) for n=[0..5] and x = [1..(floor(E^(2^n))-(floor(E^(2^(n-1))))^2+(abs(n-(1/2))/((1/2)-n)))-1]:

x^2-x+[floor(E^(2^n))-(floor(E^(2^(n-1))))^2+(abs(n-(1/2))/((1/2)-n))] 

I think that this is a very nice application of a Mills-like constant, we can obtain the whole set of Euler's "Lucky" numbers just using a single expression.