Cheers,

I changed the line:
%e A112046 It appears that every term is prime (A000040).
erroneously on %e, not %C line!
to a brief and simple proof that indeed it is so,
every term is prime:


%I A112046
%S A112046 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,
%T A112046 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,
%U A112046 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
%N A112046 a(n) = The first i >= 1, for which Jacobi symbol J(i,2n+1) is not +1 (i.e. is either 0 or -1).
%C A112046 If we instead list the first i >= 1, for which Jacobi symbol J(i,2n+1) is 0, we get A090368.
%C A112046 It is easy to see that every term is prime: Because 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.
%Y A112046 a(n) = A000040(A112049(n)) = A112050(n)+1. Bisections: A112047, A112048. Their difference: A112053. C.f. A112060, A112070.
%K A112046 nonn
%O A112046 1,1
%A A112046 Antti Karttunen (His-Firstname.His-Surname(AT)iki.fi), Aug 27 2005