A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).

0, 1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 24, 25, 26, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 50, 52, 53, 55, 56, 57, 59, 60, 62, 63, 65, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 82, 83, 84, 86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 100

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

Earliest monotonic sequence greater than 0 satisfying the condition: "a(n) + 2n is not in the sequence". - Benoit Cloitre, Mar 25 2004

Also the integer part of the hypotenuse of isosceles right triangles. The real part of these numbers is irrational. For proof see Jones and Jones.

First differences are 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, ... (A006337 with a 1 in front). - Philippe Deléham, May 29 2006

It appears that the distance between the a(n)-th triangular number and the nearest square is not greater than floor(a(n)/2). - Ralf Stephan, Sep 14 2013

These are the nonnegative integers m satisfying sin(m*Pi/r)*sin((m+1)*Pi/r) <= 0, where r = sqrt(2). In general, the Beatty sequence of an irrational number r > 1 consists of the numbers m satisfying sin(m*x)*sin((m+1)*x) <= 0, where x = Pi/r. Thus the numbers m satisfying sin(m*x)*sin((m+1)*x) > 0 form the Beatty sequence of r/(1-r). - Clark Kimberling, Aug 21 2014

For n > 0: A080764(a(n)) = 1. - Reinhard Zumkeller, Jul 03 2015

From Clark Kimberling, Oct 17 2016: (Start)

We can generate A001951 and A001952 without using sqrt(2).

First write the even positive integers in a row:

2 4 6 8 10 12 14 . . .

Then put 1 under 2 and add:

2 4 6 8 10 12 14 . . .

1

3

Next, under 4, put the least positive integer that is not yet in rows 2 and 3;

it is 2; and add:

2 4 6 8 10 12 14 . . .

1 2

3 6

Next, under the 6 in row 1, put the least positive integer not yet in rows 2 and 3;

it is 4, and add:

2 4 6 8 10 12 14 . . .

1 2 4

3 6 10

Continue in this manner. (End)

This sequence contains an infinite number of powers of 2 (proof in Crux Mathematicorum link). See A103341. - Bernard Schott, Mar 08 2019

The terms of this sequence generate the multiplicative group of positive rational numbers (observation by Stephen M. Gagola, Jr.; see References). - Allen Stenger, Aug 05 2023

a(n) is also the number of distinct straight cylinders with integer radius and height having the same surface as a sphere with radius n. - Felix Huber, Sep 20 2024

Let P(x,y) be the condition x^2 + y^2 <= n^2, then 4*a(n) is the number of integer points (x,y) such that P(x,y) is true and at least one of P(x+1,y), P(x-1,y), P(x,y+1), P(x,y-1) is false. See LINKS for examples. - Bob de Boisvilliers, May 14 2025

References

• Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
• Stephen M. Gagola Jr., Solution of Problem 12282, Am. Math. Monthly, 130 (2023), pp. 682-683.
• R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 77.
• Gareth A. Jones and J. Mary Jones, Elementary Number Theory, Springer, 1998; pp. 221-222.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
• Roland Sprague, Recreations in Mathematics, Blackie and Son, (1963).
• David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, Revised edition (1997), Entry sqrt(2), p. 18.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• L. Carlitz, Richard Scoville and Verner E. Hoggatt, Jr., Pellian representatives, Fib. Quart., 10 (1972), 449-488.
• Ian G. Connell, A generalization of Wythoff's game, Canad. Math. Bull. 2 (1959) 181-190
• Bob de Boisvilliers, ASCII circles' boundary pixels count examples from n=0 to n=10
• Ed Doolittle, Problem 19, 26th I.M.O. Finland proposed by Romania, Crux Mathematicorum, p. 70, Vol. 14, Mar. 88.
• Aviezri S. Fraenkel, How to beat your Wythoff games' opponent on three fronts, Amer. Math. Monthly, 89 (1982), 353-361 (the case a=2).
• Aviezri S. Fraenkel, On the recurrence f(m+1)= b(m)*f(m)-f(m-1) and applications, Discrete Mathematics 224 (2000), no. 1-3, pp. 273-279.
• Wen An Liu and Xiao Zhao, Adjoining to (s,t)-Wythoff's game its P-positions as moves, Discrete Applied Mathematics, Aug 27 2014; See Table 3.
• Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See pp. 17-18.
• N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
• Eric Weisstein's World of Mathematics, Beatty Sequence.
• Index entries for sequences related to Beatty sequences

Crossrefs

Complement of A001952. Equals A001952(n) - 2*n for n>0.

Equals A003151(n) - n; a bisection of A094077.

Bisections: A022842, A342281.

Cf. A022342, A026250, A080764, A103341.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021

Partial sums: A194102.

Programs

Haskell

a001951 = floor . (* sqrt 2) . fromIntegral
-- Reinhard Zumkeller, Sep 14 2014

Magma

[Floor(n*Sqrt(2)): n in [0..60]]; // Vincenzo Librandi, Oct 22 2011

Magma

[Isqrt(2*n^2):n in[0..60]]; // Jason Kimberley, Oct 28 2016

Maple

a:=n->floor(n*sqrt(2)): seq(a(n),n=0..80); # Muniru A Asiru, Mar 09 2019

Mathematica

Floor[Range[0, 72] Sqrt[2]] (* Robert G. Wilson v, Oct 17 2012 *)

Maxima

makelist(floor(n*sqrt(2)), n, 0, 100); /* Martin Ettl, Oct 17 2012 */

PARI

f(n) = for(j=1,n,print1(floor(sqrt(2*j^2))","))

PARI

a(n)=sqrtint(2*n^2) \\ Charles R Greathouse IV, Oct 19 2016

Python

from sympy import integer_nthroot
def A001951(n): return integer_nthroot(2*n**2,2)[0] # Chai Wah Wu, Mar 16 2021

Formula

a(n) = A000196(A001105(n)). - Jason Kimberley, Oct 26 2016

a(n) = floor(csc(1/(sqrt(2)*n))) for n > 0, since sqrt(2)*n < csc(1/(sqrt(2)*n)) < sqrt(2)*n + 1/(3*sqrt(2)*n) < floor(sqrt(2)*n) + 1 for n > 0. - Jianing Song, Sep 07 2021

a(n) = A194102(n) - A194102(n-1) for n > 0. - M. F. Hasler, Apr 23 2022

Extensions

More terms from David W. Wilson, Sep 20 2000