

A112046


a(n) = the least k >= 1 for which the Jacobi symbol J(k,2n+1) is not +1 (thus is either 0 or 1).


21



2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 5, 7, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 7, 5, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 7, 11, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 5, 5, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 5, 13, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 7, 5, 2, 2, 3, 3, 2, 2
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OFFSET

1,1


COMMENTS

If we instead list the least k >= 1, for which Jacobi symbol J(k,2n+1) is 0, we get A090368.
It is easy to see that every term is prime. Because the Jacobi symbol is multiplicative as J(ab,m) = J(a,m)*J(b,m) and if for every index i>=1 and < x, J(i,m)=1, then if J(x,m) is 0 or 1, x cannot be composite (say y*z, with both y and z less than x), as then either J(y,m) or J(z,m) would be nonone, which contradicts our assumption that x is the first index where nonone value appears. Thus x must be prime.


LINKS



FORMULA



PROG

(PARI) A112046(n) = for(i=1, (2*n), if((kronecker(i, (n+n+1)) < 1), return(i))); \\ Antti Karttunen, May 26 2017
(Python)
from sympy import jacobi_symbol as J
def a(n):
i=1
while True:
if J(i, 2*n + 1)!=1: return i
else: i+=1


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



