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A132222
Beatty sequence 1+2*floor(n*Pi/2), which contains infinitely many primes.
0
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
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
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Aug 14 2007
EXTENSIONS
More terms from Stefan Steinerberger, Sep 02 2007
STATUS
approved