Equivalence of Definitions of Balanced Prime

Theorem

The following definitions of a balanced prime are equivalent:

Definition 1

Let $\left({p_{n - 1}, p_n, p_{n + 1} }\right)$ be a triplet of consecutive prime numbers.

$p_n$ is a balanced prime if and only if:

$p_n = \dfrac {p_{n - 1} + p_{n + 1} } 2$

Definition 2

Let $\paren {p_{n - 1}, p_n, p_{n + 1} }$ be a triplet of consecutive prime numbers.

$p_n$ is a balanced prime if and only if:

\(\ds p_{n - 1} + d\)

\(=\)

\(\ds p_n\)

\(\ds p_{n - 1} + 2 d\)

\(=\)

\(\ds p_{n + 1}\)

for some $d \in \Z$.

Definition 3

Let $p$ be a prime number.

$p$ is a balanced prime if and only if the prime gaps either side of it are equal.

Proof

\(\ds p_n\)

\(=\)

\(\ds \dfrac {p_{n - 1} + p_{n + 1} } 2\)

Definition 1 of Balanced Prime

\(\ds \leadstoandfrom \ \ \)

\(\ds 2 p_n\)

\(=\)

\(\ds p_{n - 1} + p_{n + 1}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds p_n - p_{n - 1}\)

\(=\)

\(\ds p_{n + 1} - p_n\)

Definition 3 of Balanced Prime: equal prime gaps

\(\ds \)

\(=\)

\(\ds d\)

for some $d \in \Z$

\(\ds \leadstoandfrom \ \ \)

\(\ds p_{n - 1} + d\)

\(=\)

\(\ds p_n\)

Definition 2 of Balanced Prime

\(\ds p_{n - 1} + 2 d\)

\(=\)

\(\ds p_{n + 1}\)

$\blacksquare$