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).

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

Offset: 1

Author

Antti Karttunen, Aug 27 2005

Keywords

• nonn

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 non-one, which contradicts our assumption that x is the first index where non-one 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)

Crossrefs

One more than A112050.

Bisections: A112047, A112048, and their difference: A112053.

Cf. A000040, A053760, A053761, A090368, A112049, A112060, A112070, A268829, A286465, A286466.

Programs

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

Formula

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