A351319 a(n) = floor(n/k), where k is the nearest square to n.

1, 2, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1

Offset

1,2

Comments

For all n != 2, a(n) is 0 when less than the nearest square, A053187(n), and is 1 otherwise.

From Jon E. Schoenfield, Mar 22 2022: (Start)

After the first two terms, the sequence consists of runs of 0's and 1's, with run lengths 1,3,2,4,3,5,4,6,5,7,6,8,... = A028242.

For m >= 1, there are 2m integers k whose nearest square is m^2, namely, the m-1 integers (in the interval [m^2-m+1, m^2-1]) for which k < m^2 (hence a(k) = 0), followed by the m+1 integers (in the interval [m^2, m^2+m]) for which k >= m^2 (hence a(k) = 1). (End)

Links

Table of n, a(n) for n=1..85.

Formula

a(n) = floor(n/k), where k = round(sqrt(n))^2 = A053187(n).

a(n) = A267708(n) for n != 2.

Example

a(5) = floor(5/4) = 1.

Mathematica

Table[Floor[n/Round[Sqrt[n]]^2], {n, 100}] (* Wesley Ivan Hurt, Mar 18 2022 *)

Prog

(Python)
import math
def a(n):
    k = math.isqrt(n)
    if n - k**2 > k: k += 1
    return n // k**2;
for n in range(1, 101):
    print("{}, ".format(a(n)), end="")
(Python)
from math import isqrt
def A351319(n): return n if n <= 2 else int((k:=isqrt(n))**2+k-n+1 > 0) # Chai Wah Wu, Mar 26 2022
(PARI) a(n) = if(n==2, 2, my(r, s=sqrtint(n, &r)); r<=s); \\ Kevin Ryde, Mar 23 2022

Crossrefs

Cf. A000194, A053187 (nearest square), A028242 (run lengths).

Cf. A267708 (essentially the same).

Sequence in context: A133008 A102550 A293139 * A213725 A213726 A116929

Adjacent sequences: A351316 A351317 A351318 * A351320 A351321 A351322

Keyword

nonn,easy

Author

Joelle H. Kassir, Mar 18 2022

Status

approved