|
%I #50 Nov 08 2016 21:00:20
%S 6,9,10,12,15,18,20,24,26,27,30,36,40,42,45,48,50,52,54,60,63,66,70,
%T 72,74,75,78,80,81,84,86,90,96,99,100,102,104,105,106,108,110,111,114,
%U 117,120,126,130,132,134,135,138,140,144,148,150,153,156,159,160,162,165,166,168,170
%N Real positive integers with more than one factorization in Z[sqrt(10)].
%C 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 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 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 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 9 = 3^2 = (-1)(1 - sqrt(10))(1 + sqrt(10)), so 9 is in the sequence.
%e 10 = 2 * 5 = (sqrt(10))^2, so 10 is in the sequence.
%Y Cf. A097955, A262828.
%K nonn
%O 1,1
%A _Alonso del Arte_, Dec 23 2015
|
