cp's OEIS Frontend

Note that a(n) is always a prime q > prime(n).

For n > 1, a(n) = prime(k), where k is the smallest number such that A053760(k) = prime(n).

One could make a case for setting a(1) = 2, but a(1) = 3 seems more in keeping with the spirit of the sequence.

a(n) is the smallest odd prime q such that prime(n)^((q-1)/2) == -1 (mod q) and b^((q-1)/2) == 1 (mod q) for every natural base b < prime(n). - Thomas Ordowski, May 02 2019

As is well known, for an odd prime p, b(p) is the smallest quadratic non-residue b modulo p if and only if b(p) is the smallest base b such that b^((p-1)/2) == -1 (mod p). Note that b(n) is always a prime.

Conjecture: If 2^((n-1)/2) == -1 (mod n), then b(n) = 2, where b(n) as above. This is true for odd primes n; is it for odd composites n? If so, then all composite numbers n such that 2^((n-1)/2) == -1 (mod n) are in this sequence.

It seems that, for defined pseudoprimes n (similar to the odd primes p),

b(n) is the smallest base b such that b^((n-1)/2) == -1 (mod n), although this is not required by their definition.

Note: a "non-residue" pseudoprime n is a strong pseudoprime to base b(n); the Jacobi symbol (b(n)/n) = -1, where b(n) is the smallest non-residue modulo n; such a pseudoprime n is not a Proth number, so n = k*2^m + 1 with odd k > 2^m.

Problem: are there infinitely many such numbers?