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A132222 Beatty sequence 1+2*[n*Pi/2], which contains infinitely many primes. +0
1
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)
OFFSET

0,2

COMMENT

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.

LINKS

William D. Banks and Igor E. Shparlinski, Prime numbers with Beatty sequences, arXiv:0708.1015

FORMULA

a(n) = 1+2*[n*Pi/2] = 1+2*floor(n*Pi/2).

EXAMPLE

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.

MATHEMATICA

Table[1 + 2*Floor[n*Pi/2], {n, 0, 60}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Sep 02 2007

CROSSREFS

Cf. A019669, A130568.

Sequence in context: A087550 A047241 A086515 this_sequence A111225 A032678 A073671

Adjacent sequences: A132219 A132220 A132221 this_sequence A132223 A132224 A132225

KEYWORD

easy,nonn

AUTHOR

Jonathan Vos Post (jvospost3(AT)gmail.com), Aug 14 2007

EXTENSIONS

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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Last modified March 19 13:18 EDT 2010. Contains 173632 sequences.


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