There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers

Theorem
Let $n \in \N$ be a natural number.

Let $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ be the $n$th, $n + 1$th, $n + 2$th and $n + 3$th triangular numbers respectively.

Then it is not the case that all of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are sphenic numbers.

Proof
Let $\map \Omega n$ denote the number of prime factors of $n$ counted with multiplicity.

there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.

Thus from Closed Form for Triangular Numbers:

Recall the definition of sphenic number:

Hence:

Suppose one of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.

Then at least one of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ is divisible by $4$, and so is not sphenic.

Hence none of $n$, $n + 1$, $n + 2$, and $n + 3$ is divisible by $8$.

Thus:
 * $\map \Omega n = \map \Omega {n + 2} = \map \Omega {n + 4}$

But $n$, $n + 2$ and $n + 4$ cannot be all primes unless $n = 3$.

Thus:
 * $\map \Omega n \ge 2$

Suppose $\map \Omega n = 2$.

Then $\map \Omega {n + 1} = 2$ because $\map \Omega {n \paren {n + 1} } = 4$.

Similarly:
 * $\map \Omega {n + 2} = \map \Omega {n + 3} = \map \Omega {n + 4} = 2$

This is impossible because at least one of $n$, $n + 1$, $n + 2$, $n + 3$ and $n + 4$ is divisible by $4$.

Thus this number can only be $4$.

Thus:
 * $\map \Omega n = 3$

and:
 * $\map \Omega {n + 1} = \map \Omega {n + 3} = 1$

Thus $n + 1$ and $n + 3$ are twin primes.

Thus neither $n + 1$ nor $n + 3$ is divisible by $3$.

Thus $n + 2$ is divisible by $3$.

But $n + 2$ is not divisible by $12$.

This is because we have that $\map \Omega {n + 2} = 3$, and so it must be the case that $n + 2$ can only be $12$.

Thus $n$ and $n + 4$ must be divisible by $4$.

Note that neither $n + 1$ nor $n + 3$ cannot be divisible by $4$ because $n + 1$ and $n + 3$ are twin primes.

Hence one of $n$ and $n + 4$ will be divisible by $8$.

This contradicts our deduction that neither $n$ nor $n + 4$ can be divisible by $8$.

Also see
(Note that 406 is the smallest starting triangular number of 3 consecutive triangular numbers which are all sphenic numbers)