%I #13 Oct 19 2022 12:59:02
%S 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,
%T 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,
%U 11,11,11,11,11,11,11,23,23,17,17,17,17,17,17,17,17
%N 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.
%C This sequence gives the prime numbers associated with A357477.
%C 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.
%H Rémy Sigrist, <a href="/A357899/b357899.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = round(sqrt(A357477(n) * n)).
%e For n = 19:
%e - we have:
%e k round(sqrt(k*19)) prime?
%e - ----------------- ------
%e 1 4 No
%e 2 6 No
%e 3 8 No
%e 4 9 No
%e 5 10 No
%e 6 11 Yes
%e - so a(19) = 11.
%o (PARI) a(n) = my (p); for (k=1, oo, if (isprime(p=round(sqrt(k*n))), return (p)))
%o (Python)
%o from math import isqrt
%o from itertools import count
%o from sympy import isprime
%o 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
%Y Cf. A000194, A308052, A357477 (corresponding k's).
%K nonn
%O 1,1
%A _Rémy Sigrist_, Oct 19 2022