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Search: id:A069941
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| A069941 |
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Number of primes p such that n!<= p <= n!+n^2. |
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+0
1
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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
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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)>ln(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+ln(m)^2" since n^2<ln(n!)^2 for n>5.
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FORMULA
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a(n)=Card( k : 1<=k<=n^2 : n!+k is prime)
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PROGRAM
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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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Cf. A037151.
Sequence in context: A105590 A073750 A120204 this_sequence A002036 A090903 A022878
Adjacent sequences: A069938 A069939 A069940 this_sequence A069942 A069943 A069944
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KEYWORD
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easy,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), May 04 2002
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