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A069941 Number of primes p such that n! <= p <= n! + n^2.

%I A069941 #19 Jul 10 2015 21:46:32
%S A069941 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,
%T A069941 12,9,12,12,17,13,6,12,18,15,13,15,17,15,23,19,12,13,19,18,22,20,19,
%U A069941 16,17,19,19,23,20,18,19,23,24,19,15,19,20,26,18,24,22,24,25,24,16,23
%N A069941 Number of primes p such that n! <= p <= n! + n^2.
%C A069941 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.
%C A069941 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
%H A069941 T. D. Noe, <a href="/A069941/b069941.txt">Table of n, a(n) for n=1..300</a>
%H A069941 T. D. Noe, <a href="http://www.sspectra.com/math/A069941Theory.gif">Theoretical versus Actual</a>
%t A069941 Table[Length[Select[Range[n!,n!+n^2], PrimeQ]], {n,100}] (* _T. D. Noe_, Mar 06 2010 *)
%o A069941 (PARI) for(n=1,75,print1(sum(k=n!,n!+n^2,isprime(k)),","))
%Y A069941 Cf. A037151, A037153, A014085.
%K A069941 easy,nonn
%O A069941 1,2
%A A069941 _Benoit Cloitre_, May 04 2002