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A184774
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Primes of the form floor(k*sqrt(2)).
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21
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2, 5, 7, 11, 19, 29, 31, 41, 43, 53, 59, 67, 73, 79, 83, 89, 97, 101, 103, 107, 113, 127, 131, 137, 149, 151, 173, 179, 181, 193, 197, 199, 223, 227, 229, 233, 239, 241, 251, 257, 263, 271, 277, 281, 311, 313, 337, 347, 349, 353, 359, 367, 373, 379, 383, 397
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OFFSET
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1,1
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COMMENTS
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Let N={1,2,...}, L={floor(n*sqrt(2)): n in N} and U={2n+L(n): n in N}. Every prime is in L or U, since the union of the (disjoint) sets L and U is N.
The conjecture formerly posted here, that "if r is an irrational number and 1<r<2, then there are infinitely many primes in the set L={floor(n*r)}," is proved in I. M. Vinogradov, The Method of Trigonometrical Sums in the Theory of Numbers, (1954), page 180.
Note that every prime not in L is in U={floor(n*s}, where s=r/(r-1). That is, Beatty sequences partition the primes into two infinite classes.
The conjecture generalized: if r is a positive irrational number and h is a real number, then each of the sets {floor(n*r+h)}, {round(n*r+h)}, and {ceiling(n*r+h)} contains infinitely many primes. Can the method in Vinogradov be extended to cover these cases?
[Update regarding the conjecture from Clark Kimberling, Jan 03 2011.]
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LINKS
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FORMULA
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EXAMPLE
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The sequence L(n)=floor(n*sqrt(2)) begins with 1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19,..., which includes primes L(2)=2, L(4)=5, L(5)=7,...
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MATHEMATICA
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r=2^(1/2); s=r/(r-1);
Table[a[n], {n, 1, 120}]
t1={}; Do[If[PrimeQ[a[n]], AppendTo[t1, a[n]]], {n, 1, 600}]; t1
t2={}; Do[If[PrimeQ[a[n]], AppendTo[t2, n]], {n, 1, 600}]; t2
t3={}; Do[If[MemberQ[t1, Prime[n]], AppendTo[t3, n]], {n, 1, 300}]; t3
t4={}; Do[If[PrimeQ[b[n]], AppendTo[t4, b[n]]], {n, 1, 600}]; t4
t5={}; Do[If[PrimeQ[b[n]], AppendTo[t5, n]], {n, 1, 600}]; t5
t6={}; Do[If[MemberQ[t4, Prime[n]], AppendTo[t6, n]], {n, 1, 300}]; t6
(* the lists t1, t2, t3, t4, t5, t6 match the sequences
Select[Floor[Range[500]Sqrt[2]], PrimeQ] (* Harvey P. Dale, Jan 05 2019 *)
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PROG
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(Magma) [Floor(n*Sqrt(2)): n in [1..400] | IsPrime(Floor(n*Sqrt(2)))]; // Vincenzo Librandi, Apr 30 2015
(Python)
from math import isqrt
from itertools import count, islice
from sympy import isprime
def A184774_gen(): # generator of terms
return filter(isprime, (isqrt(k**2<<1) for k in count(1)))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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