%N Primes p such that the ring of algebraic integers of Q(sqrt(p)) does not have unique factorization.
%C It is still unknown whether there are infinitely many real, positive, squarefree d such that O_(Q(sqrt(d)) has unique factorization (or, to put it another way, the class number is 1).
%C If one only looks at small prime numbers, one could easily be tempted to think that if p is prime then O_(Q(sqrt(p)) has unique factorization.
%C By contrast, given distinct primes p and q, one could think that O_(Q(sqrt(pq)) generally does not have unique factorization, especially if p = 5.
%C It then often happens that both p and q are irreducible, and therefore pq = (sqrt(pq))^2 represents two distinct factorizations of the same number.
%C Such an obvious example of multiple distinct factorizations is obviously not available in O_(Q(sqrt(p)).
%H G. C. Greubel, <a href="/A278837/b278837.txt">Table of n, a(n) for n = 1..1000</a>
%e In Z[sqrt(79)], to pick just one example of a number having more than one distinct factorization, we verify that 3 and 5 are both irreducible, yet 15 = 3 * 5 = (-1)(8 - sqrt(79))(8 + sqrt(79)). Thus 79 is in the sequence.
%e Z[sqrt(83)] is a unique factorization domain, hence 83 is not in the sequence.
%t Select[Prime[Range], NumberFieldClassNumber[Sqrt[#]] > 1 &]
%Y Cf. A146209.
%A _Alonso del Arte_, Nov 28 2016
%E Missing term 2089 added by _Emmanuel Vantieghem_, Mar 08 2019