cp's OEIS Frontend

The next two terms are 171!+397 and 190!+409, which are too large to include. - Jud McCranie, Jul 04 2000

Conjecture: if n>=2 there are at least 3 primes p such that n!<=p<=n!+n^2 (or stronger: for n>1, a(n) > log(n)). This is stronger than the conjecture described in A037151(n). Because if n!+k is prime, k composite, k=A*B, where A and B must have, each one, at least one prime factor>n (if not: A=q*A' q<=n then n!+k is divisible by q), hence k>n^2. Also stronger (but more restrictive) than the Schinzel conjecture: "for m large enough there's at least one prime p such that m <= p <= m + log(m)^2" since n^2 < log(n!)^2 for n>5.

For the n-th term we have a(n) = pi(n!+n^2) - pi(n!), where pi(x) is the prime counting function. However, pi(n!) is difficult to compute for n>25. The Prime Number Theorem states that pi(x) and Li(x), the logarithmic integral, are asymptotically equal. Hence we can approximate a(n) by Li(n!+n^2) - Li(n!). These approximate values of a(n) are plotted as the red curve in the "Theoretical versus Actual" plot. By the way, using x/log(x) as approximation for Li(x) would change the curve by at most 1 unit. - T. D. Noe, Mar 06 2010

If n! is the average of its closest prime neighbors then the smaller prime is to be chosen (as in A051701).