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A069923
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Number of primes p such that 2^n<=p<=2^n+prime(n).
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0
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2, 2, 2, 3, 3, 3, 3, 4, 2, 5, 3, 5, 5, 4, 7, 9, 4, 5, 5, 7, 3, 4, 7, 3, 7, 6, 8, 6, 5, 8, 4, 6, 10, 3, 5, 3, 7, 6, 7, 7, 8, 6, 7, 5, 7, 5, 8, 4, 2, 7, 6, 6, 7, 3, 6, 6, 11, 6, 6, 9, 8, 8, 7, 7, 6, 6, 10, 8, 7, 10, 9, 7, 5, 5, 9, 6, 8, 11, 9, 5, 8, 6, 10, 9, 5, 9, 12, 6, 7, 4, 7, 6, 9, 8, 5, 7, 6, 7, 3, 4, 8
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| For any n>0 is there always at least one prime p such that 2^n<=p<=2^n+prime(n)? (checked until n=250 ) In this case, that would be stronger than the Schinzel conjecture : "for m >1 there's at least one prime p such that m<=p<=m+ln(m)^2" since for n >2 prime(n)<ln(2^n)^2=n^2*ln(2).
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PROG
| (PARI) for(n=1, 65, print1(sum(i=2^n, 2^n+prime(n), isprime(i)), ", "))
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CROSSREFS
| Cf. A014210.
Sequence in context: A097561 A162345 A048689 * A095840 A131343 A089051
Adjacent sequences: A069920 A069921 A069922 * A069924 A069925 A069926
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KEYWORD
| easy,nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), May 05 2002
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