A105224 Number of squares between n and 2*n inclusive.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 4

Offset

1,8

Comments

a(n)>=1 because between n and 2*n there is always at least one square.

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Formula

a(n) = A000196(2*n) - A000196(n-1).

Example

a(8)=2 because between 8 and 16 (inclusive) there are two squares: 3^2 = 9 and 4^2 = 16.

Mathematica

f[n_] := Floor[Sqrt[n]]; Table[f[2n] - f[n - 1], {n, 100}]

Prog

(PARI) first(n) = { my(res = vector(n), t = 1); res[1] = 1; for(i = 2, n, t+=(issquare(2*i) + issquare(2*i-1) - issquare(i-1)); res[i] = t ); res } \\ David A. Corneth, Jul 22 2021
(PARI) a(n) = sqrtint(2*n)-sqrtint(n-1) \\ David A. Corneth, Jul 22 2021
(Python)
from math import isqrt
def a(n): return isqrt(2*n) - isqrt(n-1)
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Jul 22 2021

Crossrefs

Cf. A000196, A105223.

Sequence in context: A331383 A201208 A006513 * A261627 A237112 A238013

Adjacent sequences: A105221 A105222 A105223 * A105225 A105226 A105227

Keyword

nonn,easy

Author

Giovanni Teofilatto, Apr 14 2005

Extensions

Edited and extended by Ray Chandler, Apr 16 2005

Status

approved