A022846 Nearest integer to n*sqrt(2).

0, 1, 3, 4, 6, 7, 8, 10, 11, 13, 14, 16, 17, 18, 20, 21, 23, 24, 25, 27, 28, 30, 31, 33, 34, 35, 37, 38, 40, 41, 42, 44, 45, 47, 48, 49, 51, 52, 54, 55, 57, 58, 59, 61, 62, 64, 65, 66, 68, 69, 71, 72, 74, 75, 76, 78, 79, 81, 82, 83, 85, 86, 88, 89, 91, 92, 93, 95, 96

Offset

0,3

Comments

Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; n^2 is in antidiagonal number a(n). Proof: n^2 is in antidiagonal m iff A000217(m-1)< n^2 <=A000217(m), where A000217(m)=m*(m+1)/2. So m = A002024(n^2) = round(n*sqrt(2)) = a(n). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Mar 07 2003

In the rectangle R(i,j), n^2 is the number in row i=A057049(n) and column j=A057050(n), so that for n >= 1, a(n) = -1 + A057049(n) + A057050(n). - Clark Kimberling, Jan 31 2011

Number of triangular numbers less than n^2. - Philippe Deléham, Mar 08 2013

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Clark Kimberling, Beatty sequences and trigonometric functions, Integers 16 (2016), #A15.

Formula

a(n) = A002024(n^2).

a(n+1) - a(n) = 1 or 2. - Philippe Deléham, Mar 08 2013

Example

n = 4, n^2 = 16; 0, 1, 3, 6, 10, 15 are triangular numbers in interval [0, 16); a(4) = 6. - Philippe Deléham, Mar 08 2013

Mathematica

Round[Sqrt[2]Range[0, 70]] (* Harvey P. Dale, Jun 18 2013 *)

Prog

(PARI) a(n)=round(n*sqrt(2))
(Magma) [Round(n*Sqrt(2)): n in [0..60]]; // Vincenzo Librandi, Oct 22 2011
(Haskell)
a022846 = round . (* sqrt 2) . fromIntegral
-- Reinhard Zumkeller, Mar 03 2014
(Python)
from math import isqrt
def A022846(n): return isqrt(n**2<<3)+1>>1 # Chai Wah Wu, Feb 10 2023

Crossrefs

Cf. A063957 (complement of this set).

Cf. A214848 (first differences), also A006338.

Sequence in context: A047299 A186159 A184578 * A083922 A039042 A007378

Adjacent sequences: A022843 A022844 A022845 * A022847 A022848 A022849

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved