cp's OEIS Frontend

A232528 Numbers n such that for all primes p where p and p-n are quadratic residues (mod 4*n), 4*p can be written as x^2 + n*y^2.

%I A232528 #33 Jun 09 2018 08:48:25
%S A232528 1,2,3,4,5,6,7,8,9,10,11,12,13,15,16,18,19,20,21,22,24,25,27,28,30,32,
%T A232528 33,35,36,37,40,42,43,45,48,51,52,57,58,60,64,67,70,72,75,78,84,85,88,
%U A232528 91,93,96,99,100,102,105,112,115,120,123,130,132,133,147,148,160,163,165,168,177,180,187,190,192,195
%N A232528 Numbers n such that for all primes p where p and p-n are quadratic residues (mod 4*n), 4*p can be written as x^2 + n*y^2.
%C A232528 Convenient numbers (A000926) are numbers n such that for all primes p where p and p-n are quadratic residues (mod 4*n), p can be written as x^2 + n*y^2, so convenient numbers are a subsequence.
%C A232528 All non-convenient numbers which are members of this sequence are either congruent to 0 (mod 4) or 3 (mod 4).
%C A232528 The equation 4*p = x^2 + n*y^2 is important because if n is a squarefree integer congruent to 3 (mod 4), then the ring of integers Q[sqrt(-n)] will be all integers of form (x/2) + (y/2)*sqrt(-n) for x and y of the same parity, whose norm is (x/2)^2 + n*(y/2)^2. If prime p = (x/2)^2 + n*(y/2)^2, then 4*p = x^2 + n*y^2.
%C A232528 Is this sequence finite?
%C A232528 Is 7392 the largest term of this sequence?
%C A232528 There are no further terms up to 10^6. - _Andrew Howroyd_, Jun 08 2018
%H A232528 V. Raman, <a href="/A232528/b232528.txt">Table of n, a(n) for n = 1..139</a>
%H A232528 Thomas R. Hagedorn, <a href="http://math.uga.edu/~pete/Hagedorn11.pdf">Primes of the form x^2+ny^2 and the geometry of (convenient) numbers</a>
%e A232528 n = 14 is not a member of this sequence because for prime p = 71, 4*p = 284 cannot be written as x^2 + 14*y^2.
%o A232528 (PARI) ok(n)=!#select(k->k<>2, quadclassunit(-n*if((-n)%4>1, 4, 1)).cyc) \\ _Andrew Howroyd_, Jun 08 2018
%Y A232528 Cf. A000926.
%K A232528 nonn
%O A232528 1,2
%A A232528 _V. Raman_, Nov 25 2013