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

Below, "omega" is

If $4$ consecutive triangular numbers are all sphenic numbers, let them be $n*(n+1)/2, (n+1)*(n+2)/2, (n+2)*(n+3)/2, (n+3)*(n+4)/2$, then we have $omega(n*(n+1)) = omega((n+1)*(n+2)) = omega((n+2)*(n+3)) * omega((n+3)*(n+4) = 4$, and none of $n, n+1, n+2, n+3, n+4$ is divisible by $8$ (or at least one of $n*(n+1)/2, (n+1)*(n+2)/2, (n+2)*(n+3)/2, (n+3)*(n+4)/2$ will be divisible by $4$ and cannot be sphenic number), thus omega(n) = omega(n+2) = omega(n+4), but n, n+2, n+4 cannot be all primes unless n=3, thus omega(n) must be >= 2, and if omega(n) = 2, then we have omega(n+1) = 2 (since omega(n*(n+1)) = 4), similarly, omega(n+2) = omega(n+3) = omega(n+4) = 2, which is impossible since at least one of n, n+1, n+2, n+3, n+4 is divisible by 4, thus this number can only be 4, thus omega(n) must be 3, and omega(n+1) = omega(n+3) = 1, i.e. 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, and n+2 cannot be divisible by 12 since omega(n+2) = 3 and if n+2 is divisible by 12 then 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 since n+1 and n+3 are twin primes), and thus one of n and n+4 will be divisible by 8, which is a contradiction!

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