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A069941
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Number of primes p such that n! <= p <= n! + n^2.
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2
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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MATHEMATICA
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Table[Length[Select[Range[n!, n!+n^2], PrimeQ]], {n, 100}] (* T. D. Noe, Mar 06 2010 *)
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PROG
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(PARI) for(n=1, 75, print1(sum(k=n!, n!+n^2, isprime(k)), ", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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