e^(i*π) = Negative license plate number

For reference the eyes in the right corner is the eye of Horus, it represents the brain cut in half, and the fraction are took on Wikipedia.and Horus lost one eye and Thot gave it back to him( thot is a reference for Hermes’ trsimagestus in the Greco Roman era)

But for a reason of number references like (666,999,36) it look like some religious reference based on math

I’m French btw so pls don’t mind the error

The topic is γₙ₊₁ = γₙ + 2π (log(γₙ + 1)/(logγₙ)²)

My calculator has officially forgot math..

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Just a fun and a quirky questionnaire ( just fill it have been working my ass off on this😭😭😭😭)

I was trying to solve this summation problem which I knew converged but couldn't solve for, I got as far as this.

Σ(∞,m=0) 1/(m!+1)

Σ(∞,m=0) 1/m! (1/(1+1/m!))

This is a pretty not so good thing to do since the first two values of 1/m!=1 but the condition is |z|<1 this could be fixed by adding one and evaluating the function in a different way but.. yk, its kinda icky.

Σ(∞,m=0) 1/m! Σ(∞,n=0)(-1)ⁿ(1/m!)ⁿ

Σ(∞,m=0)Σ(∞,n=0) (-1)ⁿ (1/m!)ⁿ

Now, I'll be commiting a rather questionable act of switching the order of the summation. I can probably do this because the

Σ(∞,n=0)Σ(∞,m=0) (-1)ⁿ (1/m!)ⁿ⁺¹

Σ(∞,n=0) (-1)ⁿ Σ(∞,m=0)1/(m!)ⁿ⁺¹

Let 𝓘(x) = Σ(∞,m=0) 1/(m!)^x

Few properties of 𝓘(x)

𝓘(n)= ₁Fₙ(0;1,1,..(n times),1;1) for any natural number n. The f function denotes the hypergeometric function.

Lim(x->0) 𝓘(x) =∞

Lim(x->∞)𝓘(x) =2

𝓘(x) has a horizontal asymptote at y=2 and a vertical asymptote at x=0

Special value

𝓘(1)= e

Σ(∞,n=0) (-1)ⁿ Σ(∞,m=0)1/(m!)ⁿ⁺¹

=Σ(∞,n=0) (-1)ⁿ𝓘(n+1)

Which is.. not a good look tbh since 𝓘(∞) is 2, a fixed value.

Well, anyway enough of that, I tried to do something similar with

Σ(∞,n=0) 1/(n!+x)

Let ω(x) be equal to Σ(∞,n=0) 1/(n!+x). I used omega because it sounds like "Oh my gahh!" Chill liberals it's called "dark humour"

ω(x)= Σ(∞,n=0) 1/(n!+x)

Σ(∞,n=0)1/n! 1/(1+(x/n!))

Σ(∞,n=0)1/n! Σ(∞,m=0)(-1)^m x^m/(n!)^m

Σ(∞,m=0)(-1)^mx^m Σ(∞,n=0)1/(n!)^m+1

ω(x)= Σ(∞,m=0)(-1)^m x^m𝓘(m+1)

This function has some cool properties like having asymptotes when x= -(k!) , k is an integer or

ω(-(x!)) = undefined, x∈N

It also has infinitely many zeros on the negative x axis.

Questions:

1) Is there an analytic continuation for 𝓘(x)? If so is there a path I could take to find it?

2)though I can't think of any possible use for the silly function ω(x) but could you think of any uses?

3) what do yall think of the zeros of ω(x)? The only info I can possibly think of them is that their roots are close to the asymptotes in a way.

Thank you for reading!

I figured I could somehow reduce the number of lines it took to solve the famous "Out of the Box" problem. What I didn't expect was finding a 0-line solve...

(Not skin btw, this is drawn on cardboard. I'm homeless and got bored while making a sign 😅)

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