N-Th Numbers — sourcepool

Why only prime numbers? Let’s extend the definition!

$a \in \mathbb{N}$ is an $n$-th number $\iff$ $a$ has $n+1$ divisors

Recall that, given $\{p_i\}$ the sequence of prime numbers (i.e. $\{2, 3, 5, \dots\}$), for every natural $a > 0$ there is a unique sequence $\{\alpha_i\}$ s.t. $a = \prod_i p_i^{\alpha_i}$, the $a$ factorization.

Then $a > 0$ has $n$ divisors $\iff$ $n = \prod_i (\alpha_i + 1)$

for there are $\alpha_i + 1$ independent choices of the exponents

$0 \le \delta_i \le \alpha_i$ of each divisor $d = \prod_i p_i^{\delta_i}$ of $a$.

Recall also that for $a = 1$, $\alpha_i = 0$ since $\prod_i p_i^0 = 1$.

- 0 is an $\infty$-th number (all numbers divide zero)

- 1 is the only zeroth number

- prime ($1$-th) numbers are classical prime numbers

- second numbers are the squares of prime numbers

- in general, $n+1$ is prime $\implies$ $n$-th numbers are $p_i^n$

- $a$ is a square $\iff$ $a$ is a $2n$-th number

The classical definition of prime numbers (divisible only by 1 and themselves) includes 1,

so 1 must be made an exception.

With this new definition it is clear that 1 is not prime, because it is zeroth.

Proofs

1 is the only zeroth number

$a$ is zeroth ($0$-th) $\iff$

$a$ has 1 divisor $\iff$

$\prod_i (\alpha_i + 1) = 1$ $\iff$

$\alpha_i + 1 = 1 \quad \forall\ i$ $\iff$

$\alpha_i = 0 \quad \forall\ i$ $\iff$

$a = \prod_i p_i^0 = 1$

prime numbers are classical prime numbers

$a$ is prime ($1$-th) $\iff$

$a$ has 2 divisors $\iff$

$\prod_i (\alpha_i + 1) = 2$ $\iff$

$\exists\ j$ s.t. $\alpha_j + 1 = 2$ and $\alpha_i + 1 = 1 \quad \forall\ i \neq j$ $\iff$

$\alpha_j = 1$ and $\alpha_i = 0 \quad \forall\ i \neq j$ $\iff$

$a = p_j\prod_{i \neq j} p_i^0 = p_j$

second numbers are the squares of prime numbers

$a$ is second ($2$-th) $\iff$

$a$ has 3 divisors $\iff$

$\prod_i (\alpha_i + 1) = 3$ $\iff$

$\exists\ j$ s.t. $\alpha_j + 1 = 3$ and $\alpha_i + 1 = 1 \quad \forall\ i \neq j$ $\iff$

$\alpha_j = 2$ and $\alpha_i = 0 \quad \forall\ i \neq j$ $\iff$

$a = p_j^2\prod_{i \neq j} p_i^0 = p_j^2$

$n+1$ is prime $\implies$ $n$-th numbers are $p_i^n$

$a$ is $n$-th $\iff$

$a$ has $n+1$ divisors $\iff$

$\prod_i (\alpha_i + 1) = n+1$ $\iff$ (being $n+1$ prime)

$\exists\ j$ s.t. $\alpha_j + 1 = n+1$ and $\alpha_i + 1 = 1 \quad \forall\ i \neq j$ $\iff$

$\exists\ j$ s.t. $\alpha_j = n$ and $\alpha_i = 0 \quad \forall\ i \neq j$ $\iff$

$a = p_j^n\prod_{i \neq j} p_i^0 = p_j^n$

$a$ is a square $\iff$ $a$ is a $2n$-th number

$a$ is a square $\iff$

$\exists\ b$ s.t. $a = b^2$ $\iff$

$\prod_i p_i^{\alpha_i} = \prod_i p_i^{2\beta_i}$ $\iff$

$\{\alpha_i\} = \{2\beta_i\}$ $\iff$

$a$ has $\prod_i (2 \beta_i + 1)$ divisors $\iff$

$a$ is a ($\prod_i (2 \beta_i + 1) - 1$)-th number

and $\prod_i (2 \beta_i + 1) - 1 = 2n$ $\iff$

$\prod_i (2 \beta_i + 1) - 1$ is even $\iff$

$\prod_i (2 \beta_i + 1)$ is odd $\iff$

all $2 \beta_i + 1$ are odd $\iff$

all $2 \beta_i$ are even, which is indeed true