If you trap both of them in the same room - what comes out of it?

im not sure, all i can say is [math]{\color{Red}{IT\ WOULD\ BE\ EXTREMELY\ PAINFUL}}[/math]

youtu.be/rGgoFmYiiM8?t=1h19m20s

Skip to 1h19m20s if not there already.

they are great thinkers

n=3 doesn't work. Try something that's not disproven in the first few cases next time.

What did he mean by this?

[math] \lfloor e^3 \rfloor = 20 [/math]

but floor(exp(4)) != floor(exp(3))floor(exp(1))

so I don't see how anything is disproven

they cancel eachother out

He provided a counter example fuckboi

a solution to the barnett conjecture

Right but it doesn't need to be true for n = 3 for it to still be true for infinitely many integers

To illuminate this subject

n is even for infinitely many integers n

You can tell that it's true despite not being true for n = 3

>n is even for infinitely many integers n
When was n being even ever stated?

>When was n being even ever stated?
Right there, in the post
>n is even

Well then that's just a poorly-worded, already proven conjecture, since it relies on the existence of infinitely many prime numbers, which there are.

What makes you say that it's already proven?

floor(e^n) doesn't saturate the natural numbers

This is either some 5th dimensional shitposting or top-tier autism.

I think what's happening is one person doesn't understand that a single counter example doesn't disprove something being true infinitely many times.

It doesn't have to saturate the natural numbers for there to be infinitely many of them. There are infinitely many odd numbers, but they don't contain all natural numbers.

It's always been frustrating to me that the dumbest people on Veeky Forums seem to be attracted to Veeky Forums.

Right but how do you know that it doesn't stop being prime at some point?

Since there are infinitely many inputs, there are infinitely many results, hence infinitely many prime results.

>Since there are infinitely many inputs, there are infinitely many results, hence infinitely many prime results.
How does infinitely many prime results follow?

I mean the sequence 2n where n is a natural number has "infinitely many inputs" and "infinitely many results", yet one and only one "prime result".

The amount of people who understand inter-universal wingdings theory will be doubled

Because e^n is an exponential function, where the base is not an integer. In your example, '2n' could be replaced with 3n, 10n, 2^n, 69^n, etc. and still hold true. But all bets are off with e^n

You have to prove that there are infinite primes... that's not known

>But all bets are off with e^n
Surely, including the bet that the floor composition has infinitely many primes for integer n.

Can you prove to me that it's prime infinitely many times?

(You)

>a single counter example doesn't disprove something being true infinitely many times.
There are only infinitely many natural numbers. A counterexample means that there are less cases where it's true than there are natural numbers.

number of cases in which it is true = A
number of natural numbers = B
infinity = C
A < B
C = B
A < C
number of cases in which it is true < infinity

Q.E.D.

If you have any further doubts on the matter, I refer you to the extensive literature on the half-infinite set of odd natural numbers.

You must be shitposting at this point.
en.wikipedia.org/wiki/Euclid's_theorem

Nah. But you can bet that floor(e^n) acts like a semi-random integer generator.

I'm not sure if this whole mathematics thing is for you.

>But you can bet that floor(e^n) acts like a semi-random integer generator.
Personally I would bet that a monotone increasing function would be a pretty poor random number generator.

That's not an argument. Either back up what you're saying or shut up, insults don't win you anything.

A poor one, sure, but one nonetheless.

In the same sense that f(x) = 1 is a random number generator, sure.

Lighten up, Francis.

No. There is no change in that function.

[math] \frac{d}{dx}[1] = 0 [/math]

[math] \frac{d}{dx}[e^x] \ne 0 [/math]

Dude it's a line LMAO

Okay.

>No. There is no change in that function.
Try putting floor(e^n) through the diehard RNG tests.

Every n'th derivative is greater than or equal to 0 for both functions.

I've been lurking this thread trying to figure out if there's some extremely good trolling here or just extremely stupid people here. In the mean time I'm enjoying playing with this fun little [math]\lfloor e^n \rfloor[/math] problem though.

I think it might be tractable, a few ideas to consider throwing at it are:

Expand the binomial e^n=(2+.7)^n and try to wiggle something out of the fact that powers of numbers less than 1 go to zero.

All primes greater than 3 are 1 away from a multiple of 6.

Dirichlet's theorem says there are infinitely many values of n that make an+b into a prime number when gcd(a,b)=1.

I'm not even sure if this function returns an odd number infinitely many times. Of course it's plausible that the even and odd values occur just as often, but I have no idea how to prove it.

[math]\frac{d}{du} 2u^2[/math]

GUYSLOOKIT!!!
[math](\pi^2)' = 2 \pi[/math]

No senpai, pi is a constant and as such it's differential is 0 as is the differential of it squared. Unless you define a function where pi is a variable, your statement is wrong

Is this a meme? I can't make sense out of this.

>If you trap both of them in the same room - what comes out of it?

Two people who can't speak the same language.

>Two people who can't speak the same language.

Mai niiga Mochizuki-san cannu supiku da engurishi beri naisuri.

Riemannian Arithmetic Geometry

I really doubt it's solvable, considering the fact that 'there are infinitely many primes in n^2 + 1' is open problem
The worst part is how you relate [math]e [/math] to any number-theoretic properties.
Perhaps floor(a^n) will be prime infinitely often if a is transcendental/irrational/not an integer?

>qed
desu***
now this is what i call shitposting.

the natural numbers are not the only infinite set
an infinite set contained by the natural numbers is not some fraction of infinity...
furthermore the real numbers or any other set of numbers containing the natural numbers is not some multiple of infinity (to follow this nonsense to its logical end)

Only one.

>autism

if a != b then a^n-b^n->infinity
So whatever approximation of e you take this ain't gonna work

Is that true that {q^n "mod 2"} is dense in [0..2) for a transcendent q?