

A069941


Number of primes p such that n! <= p <= n! + n^2.


2



1, 3, 3, 3, 4, 5, 5, 4, 7, 7, 9, 6, 5, 8, 4, 9, 10, 14, 8, 16, 14, 14, 7, 6, 16, 12, 12, 15, 13, 12, 9, 12, 12, 17, 13, 6, 12, 18, 15, 13, 15, 17, 15, 23, 19, 12, 13, 19, 18, 22, 20, 19, 16, 17, 19, 19, 23, 20, 18, 19, 23, 24, 19, 15, 19, 20, 26, 18, 24, 22, 24, 25, 24, 16, 23
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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 nth 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


LINKS



MATHEMATICA

Table[Length[Select[Range[n!, n!+n^2], PrimeQ]], {n, 100}] (* T. D. Noe, Mar 06 2010 *)


PROG

(PARI) for(n=1, 75, print1(sum(k=n!, n!+n^2, isprime(k)), ", "))


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



