Easy math problem

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>>8685568
Your conjecture holds until 8
Very unimpressive
back to work, mochi-san

>>8685568
This is fucking retarded

>>8685573
you don't even need to go up to 8
it fails at 5 and if you don't want to do computations it's immediate that it fails for 6

*For what positive integers n, we have that floor(n^2/3) is a prime number, sorry.

>>8685616
>implying 2 and 3 are not prime numbers
u w0t m8

we're talking about the cube root of n squared, not n squared divided by 3

>>8685652
>we're talking about the cube root of n squared, not n squared divided by 3

I see no parenthesis there and from this post sequence I assume you are not the OP so why don't you shut the fuck up.

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For positive integers n, the nth prime number is prime.

> I win everytime

>>8685568
1 is not a prime

for positive numbers x, we have that f(x)=5 yields a prime number

>>8685568
100/3
33
11*3
WEW lad

I tried it for x=2, 3, 4.

By induction it's true.

>>8685628
>*For what positive integers n, we have that floor(n^2/3) is a prime number, sorry.

3 you fucking idiot.

3^2/3 = 2.08
⌊2.08⌋ = 2

[math]\left \lfloor {x^{\frac{2}{3}}} \right \rfloor\ \text{is prime} \implies x^{\frac{2}{3}} \in\bigcup\limits_{p\ \text{prime}}[p,p+1)\\ \implies x \in \bigcup\limits_{p\ \text{prime}}[p^\frac{3}{2},(p+1)^\frac{3}{2}) [/math]

>>8685661
I don't get it.

>>8686064
Never mind.

>>8685674
he did it
he solved the primes
noble piece price

>>8685871
The way he wrote it, it's actually [math]\frac{x^2}{3}[/math] and not [math]x^{\frac{2}{3}}[/math]