cp's OEIS Frontend

A262362 Real positive integers with more than one factorization in Z[sqrt(10)].

%I A262362 #50 Nov 08 2016 21:00:20
%S A262362 6,9,10,12,15,18,20,24,26,27,30,36,40,42,45,48,50,52,54,60,63,66,70,
%T A262362 72,74,75,78,80,81,84,86,90,96,99,100,102,104,105,106,108,110,111,114,
%U A262362 117,120,126,130,132,134,135,138,140,144,148,150,153,156,159,160,162,165,166,168,170
%N A262362 Real positive integers with more than one factorization in Z[sqrt(10)].
%C A262362 To count as distinct from another factorization, a factorization must not be derived from the other by multiplication by units. For example, (4 - sqrt(10))(4 + sqrt(10)) is not distinct from (-1)(2 - sqrt(10))(2 + sqrt(10)) as a factorization of 6 because -3 - sqrt(10) is a unit and (2 - sqrt(10))(-3 - sqrt(10)) = 4 + sqrt(10).
%C A262362 Given a number p that is prime in Z, if x^2 == 10 mod p has solutions in Z, then some multiples of p are in this sequence. If x is the smallest solution, then x^2 - 10 gives the smallest multiple of p in this sequence not divisible by any prior term. For example, 6^2 == 10 mod 13, and 26 = 2 * 13 = (6 - sqrt(10))(6 + sqrt(10)).
%C A262362 If a number is in this sequence, then so are all its real positive integer multiples. The negative multiples also have more than one factorization, but of course one has to remember to put in the -1 as needed. Since Z[sqrt(10)] has units of norm -1, it is then possible to "shop" the units to include or exclude -1 from the factorization.
%C A262362 Z[sqrt(10)] has class number 2. This means that while a number may have more than one factorization, all factorizations have the same number of nonunit irreducible factors. If one factorization seems to have fewer factors, then it is an incomplete factorization.
%e A262362 9 = 3^2 = (-1)(1 - sqrt(10))(1 + sqrt(10)), so 9 is in the sequence.
%e A262362 10 = 2 * 5 = (sqrt(10))^2, so 10 is in the sequence.
%Y A262362 Cf. A097955, A262828.
%K A262362 nonn
%O A262362 1,1
%A A262362 _Alonso del Arte_, Dec 23 2015