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A278837 Primes p such that the ring of algebraic integers of Q(sqrt(p)) does not have unique factorization. 1


%S 79,223,229,257,359,401,439,443,499,577,659,727,733,761,839,1009,1087,

%T 1091,1093,1129,1171,1223,1229,1297,1327,1367,1373,1429,1489,1523,

%U 1567,1601,1627,1787,1811,1847,1901,1907,1987,2027,2029,2081,2089,2099,2143,2153,2207,2213,2251,2399,2459,2467

%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[100]], NumberFieldClassNumber[Sqrt[#]] > 1 &]

%Y Cf. A146209.

%K nonn

%O 1,1

%A _Alonso del Arte_, Nov 28 2016

%E Missing term 2089 added by _Emmanuel Vantieghem_, Mar 08 2019

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Last modified January 20 16:04 EST 2022. Contains 350472 sequences. (Running on oeis4.)