I originally thought this question was too basic for MathOverflow, so I tried Math.StackExchange, but no one there seemed to have a solution. Here goes.

How many consecutive composite integers follow `k!+1`

? I wrote a program in Mathematica to compute explicit answers for the first 300, but there doesn't seem to be much of a pattern. The results of that are here. This is a problem in Underwood Dudley's *Elementary Number Theory* (section 23.2 part (b)), but for the LIFE of me, I can't figure it out! My initial thought was this: Let `m`

be the smallest prime such that `k<m`

. Then any number of the form `k!+i`

, where `1<i<m`

will clearly be composite. Hence, there are at least `m-2`

composite numbers which follow `k!+1`

. However, past that it seems hard to say. For example, `11!+13`

factorizes as `199*200587`

, which seems to be unpredictable behavior. I'm leaning toward thinking this was a misprint in the book and it's a much harder problem than something that should be in *Elementary Number Theory*. Perhaps I'm overlooking a simple, elegant solution.

The other thing is, if there *were* a closed-form solution, then we would have an easy formula for calculating arbitrarily large primes. Take the largest known prime `p`

and suppose we determine that `q`

composite numbers follow `p!+1`

. Then `p!+2+q`

is prime.

`m-2`

, where`m`

is as I defined it (the least prime greater than or equal to`k`

). – Willy Feb 2 '11 at 9:08