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Search: id:A061265
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| A061265 |
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Number of squares between n-th prime and (n+1)st prime. |
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+0
5
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| 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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If n-th prime is a member of A053001 then a(n) is at least 1. If not, then a(n) = 0.
Legendre's conjecture (still open) that there is always a prime between n^2 and (n+1)^2 is equivalent to conjecturing that a(n)<=1 for all n. - Vladeta Jovovic (vladeta(AT)eunet.rs), May 01 2003
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LINKS
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Harry J. Smith, Table of n, a(n) for n=1,...,2000
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FORMULA
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a(n) = floor(sqrt(prime(n+1)))-floor(sqrt(prime(n))). - Vladeta Jovovic (vladeta(AT)eunet.rs), May 01 2003
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EXAMPLE
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a(3) = 0 as there is no square between 5, the third prime and 7, the fourth prime. a(4) = 1, as there is a square '9' between the 4th prime 7 and the 5th prime 11.
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PROGRAM
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(PARI) { n=0; q=2; forprime (p=3, prime(2001), write("b061265.txt", n++, " ", floor(sqrt(p))-floor(sqrt(q))); q=p ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jul 20 2009]
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CROSSREFS
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Cf. A053001.
Cf. A038107.
Sequence in context: A082848 A141743 A112416 this_sequence A125122 A000035 A131734
Adjacent sequences: A061262 A061263 A061264 this_sequence A061266 A061267 A061268
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KEYWORD
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nonn,base
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 24 2001
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EXTENSIONS
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Extended by Patrick De Geest (pdg(AT)worldofnumbers.com), Jun 05 2001.
OFFSET changed from 0,1 to 1,1 by Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jul 20 2009
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