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Search: id:A132222
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| A132222 |
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Beatty sequence 1+2*[n*Pi/2], which contains infinitely many primes. |
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1
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| 1, 3, 7, 9, 13, 15, 19, 21, 25, 29, 31, 35, 37, 41, 43, 47, 51, 53, 57, 59, 63, 65, 69, 73, 75, 79, 81, 85, 87, 91, 95, 97, 101, 103, 107, 109, 113, 117, 119, 123, 125, 129, 131, 135, 139, 141, 145, 147, 151, 153, 157, 161, 163, 167, 169, 173, 175, 179, 183, 185, 189
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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The primes in this entirely odd sequence begin 3, 7, 13, 19, 29, 31, 37, 41, 43, 47, 53, 59, 73, 79, 97, 101. By the theorems in Banks, there are an infinite number of primes in this sequence.
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LINKS
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William D. Banks and Igor E. Shparlinski, Prime numbers with Beatty sequences, arXiv:0708.1015
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FORMULA
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a(n) = 1+2*[n*Pi/2] = 1+2*floor(n*Pi/2).
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EXAMPLE
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a(0) = 1 because 1 + 2*[0*Pi] = 1 + 2*0 = 1 + 0 = 1.
a(1) = 3 because 1 + 2*[1*Pi/2] = 1 + 2*[1.5707963267948966192313216916] = 1 + 2*1 = 3.
a(2) = 7 because 1 + 2*[2*Pi/2] = 1 + 2*[3.1415926535897932384626433832] = 1 + 2*3 = 7.
a(3) = 9 because 1 + 2*[3*Pi/2] = 1 + 2*[4.7123889803846898576939650749] = 1 + 2*4 = 9.
a(4) = 13 because 1 + 2*[4*Pi/2] = 1 + 2*[6.2831853071795864769252867665] = 1 + 2*6 = 13.
a(5) = 15 because 1 + 2*[5*Pi/2] = 1 + 2*[7.8539816339744830961566084581] = 1 + 2*7 = 15.
a(7) = 21 because 1 + 2*[7*Pi/2] = 1 + 2*[10.995574287564276334619251841] = 1 + 2*10 = 21.
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MATHEMATICA
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Table[1 + 2*Floor[n*Pi/2], {n, 0, 60}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Sep 02 2007
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CROSSREFS
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Cf. A019669, A130568.
Sequence in context: A087550 A047241 A086515 this_sequence A111225 A032678 A073671
Adjacent sequences: A132219 A132220 A132221 this_sequence A132223 A132224 A132225
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KEYWORD
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easy,nonn,new
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AUTHOR
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Jonathan Vos Post (jvospost3(AT)gmail.com), Aug 14 2007
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EXTENSIONS
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More terms from Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Sep 02 2007
Replaced arxiv URL by non-cached version - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 30 2009
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