A184774 - OEIS (original) (raw)

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

COMMENTS

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.]

EXAMPLE

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,...

MATHEMATICA

r=2^(1/2); s=r/(r-1);

a[n_]:=Floor [n*r]; (* A001951 *)

b[n_]:=Floor [n*s]; (* A001952 *)

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 *)

PROG

(PARI) is(n)=my(k=sqrtint(n^2\2)+1); sqrtint(2*k^2)==n && isprime(n) \\ Charles R Greathouse IV, Apr 29 2015

(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)))