Consecutive a-smooth numbers I am looking for two large numbers $n, n+1$ such that both are $7$-smooth numbers. The two largest pairs I found are $2400, 2401$ and $4374, 4375$. Can anyone find a larger pair if it exists?
Second, are there infinite pairs of $7$-smooth numbers: $n, n+1$? 
Thanks for someone that could solve this. 
 A: I googled these things and found Størmer's theorem:
https://en.wikipedia.org/wiki/St%C3%B8rmer's_theorem
Quote:

It follows from the Thue–Siegel–Roth theorem that there are only a finite number of pairs of this type, but Størmer gave a procedure for finding them all.


A computer check up to 60 000 reveals no such pair greater then yours.
Edit:
The link http://oeis.org/A117581 provided by ccorn says there is no such number greater then yours.
A: Although this does not directly answer your question, I thought it may be of interest to you. I made this formula a while back but never had even a possible purpose for it until your question. The number of 7-smooth numbers $\le n$ is $\displaystyle\sum_{j=1}^{n}\lfloor\frac{210^j}{j}\rfloor-\lfloor\frac{210^{j}-1}{j}\rfloor.$ This may help, though I'm not sure how efficiently.
In general, for K being the greatest prime factor of a primorial number $\#p,$ the number of K-smooth numbers $\le n$ is given by $\displaystyle\sum_{j=1}^{n}\lfloor\frac{\#p^j}{j}\rfloor-\lfloor\frac{\#p^{j}-1}{j}\rfloor.$
The formula is a variation of another formula I made of this form $\displaystyle\sum_{j=1}^{n}\lfloor\frac{n^j}{j}\rfloor-\lfloor\frac{n^{j}-1}{j}\rfloor,$ which gives the number of numbers $k\le n$ whose set of prime factors is a subset of the set of prime factors of n. This works because $\displaystyle\lfloor\frac{n^j}{j}\rfloor-\lfloor\frac{n^{j}-1}{j}\rfloor$ is either $1$ or $0$ depending on whether or not $j|n^j$ or does not respectively, which can only occur if the set of factors of $j$ are a subset of $n$'s. Summing this gives that function.
I'm sorry I don't have a direct answer, but I thought the formula may be more useful in your hands and it seems your question was already answered.
