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 A278837 Primes p such that the ring of algebraic integers of Q(sqrt(p)) does not have unique factorization. 1
 79, 223, 229, 257, 359, 401, 439, 443, 499, 577, 659, 727, 733, 761, 839, 1009, 1087, 1091, 1093, 1129, 1171, 1223, 1229, 1297, 1327, 1367, 1373, 1429, 1489, 1523, 1567, 1601, 1627, 1787, 1811, 1847, 1901, 1907, 1987, 2027, 2029, 2081, 2089, 2099, 2143, 2153, 2207, 2213, 2251, 2399, 2459, 2467 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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). 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. 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. 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. Such an obvious example of multiple distinct factorizations is obviously not available in O_(Q(sqrt(p)). LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 EXAMPLE 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. Z[sqrt(83)] is a unique factorization domain, hence 83 is not in the sequence. MATHEMATICA Select[Prime[Range[100]], NumberFieldClassNumber[Sqrt[#]] > 1 &] CROSSREFS Cf. A146209. Sequence in context: A142747 A142198 A089686 * A260335 A257933 A258098 Adjacent sequences:  A278834 A278835 A278836 * A278838 A278839 A278840 KEYWORD nonn AUTHOR Alonso del Arte, Nov 28 2016 EXTENSIONS Missing term 2089 added by Emmanuel Vantieghem, Mar 08 2019 STATUS approved

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Last modified November 30 20:17 EST 2021. Contains 349425 sequences. (Running on oeis4.)