

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

Indranil Ghosh and A.H.M. Smeets, Table of n, a(n) for n = 1..20000 (first 1000 terms from Indranil Ghosh)


FORMULA

a(n) = A112050(n) + 1 = A000040(A112049(n)).


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
print([a(n) for n in range(1, 103)]) # Indranil Ghosh, May 11 2017


CROSSREFS

One more than A112050.
Bisections: A112047, A112048, and their difference: A112053.
Cf. A000040, A053760, A053761, A090368, A112049, A112060, A112070, A268829, A286465, A286466.
Sequence in context: A234972 A130326 A059906 * A076902 A287271 A290884
Adjacent sequences: A112043 A112044 A112045 * A112047 A112048 A112049


KEYWORD

nonn


AUTHOR

Antti Karttunen, Aug 27 2005


STATUS

approved



