A357899 Let k be the smallest k such that the square root of k*n rounds to a prime number; a(n) is this prime number.

2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 5, 5, 5, 7, 7, 7, 11, 11, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 11, 11, 11, 13, 13, 13, 11, 11, 11, 11, 11, 11, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 23, 23, 17, 17, 17, 17, 17, 17, 17, 17

Offset

1,1

Comments

This sequence gives the prime numbers associated with A357477.

The sequence is well defined as for any prime number p >= n/2, A000194 contains n or more consecutive p's, and so a(n) <= p.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

a(n) = round(sqrt(A357477(n) * n)).

Example

For n = 19:
- we have:
      k  round(sqrt(k*19))  prime?
      -  -----------------  ------
      1                  4  No
      2                  6  No
      3                  8  No
      4                  9  No
      5                 10  No
      6                 11  Yes
- so a(19) = 11.

Prog

(PARI) a(n) = my (p); for (k=1, oo, if (isprime(p=round(sqrt(k*n))), return (p)))
(Python)
from math import isqrt
from itertools import count
from sympy import isprime
def A357899(n): return next(filter(isprime, ((m:=isqrt(k*n))+ int((k*n-m*(m+1)<<2)>=1) for k in count(1)))) # Chai Wah Wu, Oct 19 2022

Crossrefs

Cf. A000194, A308052, A357477 (corresponding k's).

Sequence in context: A279757 A251550 A279221 * A247781 A004052 A051742

Adjacent sequences: A357896 A357897 A357898 * A357900 A357901 A357902

Keyword

nonn

Author

Rémy Sigrist, Oct 19 2022

Status

approved