On certain triplets of consecutive integers [closed]

Question

While completelely factorizing integers, my student Luciana noticed that the canonical prime factorization of the three consecutive numbers 81=3^4, 82=2x41, and 83=83, use numbers which are all different: 2,3,4,41, and 83. She would like to know the largest possible triplet of such numbers. I told her it was likely that such triplets occur infinitely often, but a matter difficult to prove.

"What then is the largest such triplet you can find?",-she asked. I told her I would pass the question to someone abler than me.

Answer 1

Going with the twin primes approach, I tried the largest twin primes found to date, ($2996863034895 · 2^{1290000} \pm 1$, but $2996863034895$ factorizes to $3^2×5×18583×3583757$ which means $2$ would be both a base and an exponent. I continued this strategy.

The second largest twin primes found to date are $3756801695685 · 2^{666669} \pm 1$. To try the same strategy again, $3756801695685$ factorizes to $3×5×43×347×16785299$. If we allow multiple exponents of 1, then this means the values can be represented the following way: $3756801695685 · 2^{666669} - 1 = $ ($200,700$ decimal digits I won't write out, but it is prime, and thus has a single factor to the power of $1$), $3756801695685 · 2^{666669}=2^{666669}×3×5×43×347×16785299$, and $3756801695685 · 2^{666669} + 1 = $ ($200,700$ decimal digits I again won't write out, but you get the deal). Therefore, all those factors are unique, and so are the non-$1$ exponents.

There are very likely infinitely many larger triplets. This is the largest I wanted to take the time to find.