Joelle H. Kassir - cp's OEIS frontend

User: Joelle H. Kassir

Joelle H. Kassir has authored 2 sequences.

A353024 Largest k such that A007504(k) <= n^2.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 45, 46, 47, 47

Offset: 1

Joelle H. Kassir, Apr 17 2022

nonn

Cf. A000290, A007504, A337769, A350174, A033997.

first(N)=my(v=vector(N),s,k,n=1,n2=1); forprime(p=2,, s+=p; k++; while(s>n2, v[n]=k-1; if(n++>N, return(v)); n2=n^2)) \\ Charles R Greathouse IV, Apr 18 2022

a(n)=my(n2=n^2,s,k); forprime(p=2,, s+=p; k++; if(s>n2, return(k-1))) \\ Charles R Greathouse IV, Apr 18 2022

from sympy import prime
def a(n):
    k = 1
    total = 0
    while True:
        total += prime(k)
        if total > n**2:
            break
        k += 1
    return k-1

a(n) = A337769(n) - 1.
a(n) ~ sqrt(2)*n/sqrt(log n). - Charles R Greathouse IV, Apr 18 2022
a(n) = A350174(n^2). - Kevin Ryde, Apr 19 2022

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

Original entry on oeis.org

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

Joelle H. Kassir, Mar 18 2022

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)

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

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

Cf. A267708 (essentially the same).

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

a(n) = if(n==2,2, my(r,s=sqrtint(n,&r)); r<=s); \\ Kevin Ryde, Mar 23 2022

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="")

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

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

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

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User: Joelle H. Kassir

A353024 Largest k such that A007504(k) <= n^2.

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

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