A132222 Beatty sequence 1+2*floor(n*Pi/2), which contains infinitely many primes.

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

Offset

0,2

Comments

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

Table of n, a(n) for n=0..60.

William D. Banks and Igor E. Shparlinski, Prime numbers with Beatty sequences, arXiv:0708.1015 [math.NT], 2007.

Formula

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

a(n) = 1 + 2*A140758(n). - L. Edson Jeffery, Mar 16 2013

Example

a(0) = 1 because 1 + 2*floor(0*Pi) = 1 + 2*0 = 1 + 0 = 1.
a(1) = 3 because 1 + 2*floor(1*Pi/2) = 1 + 2*floor(1.5707963) = 1 + 2*1 = 3.
a(2) = 7 because 1 + 2*floor(2*Pi/2) = 1 + 2*floor(3.1415926) = 1 + 2*3 = 7.
a(3) = 9 because 1 + 2*floor(3*Pi/2) = 1 + 2*floor(4.7123889) = 1 + 2*4 = 9.
a(4) = 13 because 1 + 2*floor(4*Pi/2) = 1 + 2*floor(6.2831853) = 1 + 2*6 = 13.
a(5) = 15 because 1 + 2*floor(5*Pi/2) = 1 + 2*floor(7.8539816) = 1 + 2*7 = 15.
a(7) = 21 because 1 + 2*floor(7*Pi/2) = 1 + 2*floor(10.995574) = 1 + 2*10 = 21.

Mathematica

Table[1 + 2*Floor[n*Pi/2], {n, 0, 60}] (* Stefan Steinerberger, Sep 02 2007 *)

Crossrefs

Cf. A019669, A130568, A140758.

Sequence in context: A356138 A047241 A086515 * A340933 A320634 A330122

Adjacent sequences: A132219 A132220 A132221 * A132223 A132224 A132225

Keyword

easy,nonn

Author

Jonathan Vos Post, Aug 14 2007

Extensions

More terms from Stefan Steinerberger, Sep 02 2007

Status

approved