# On certain triplets of consecutive integers [closed]

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.

• So the question is "what is the largest triplet you can find"? How is this not "subjectively correct answers", then?
– Deusovi
Jul 16, 2020 at 18:13
• The twin-primes conjecture immediately implies there are infinitely many such triplets. Jul 16, 2020 at 18:21
• @msh210 Not so sure. The number between the two primes might have, say, repeated exponents. Jul 16, 2020 at 18:54
• Oh, sorry, didn't realize you were including the exponents. But then your example is no good: it has repeated exponents of 1. Jul 16, 2020 at 18:56
• @OP I don't think this is an interesting question. It most likely is unsolved, and finding the largest such triplet is likely a game of who has the most computing power. Jul 16, 2020 at 20:35

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.