A322781 - OEIS

%I #18 Jan 12 2019 02:31:52

%S 65,85,185,265,365,481,485,493,533,565,629,685,697,785,865,949,965,

%T 985,1037,1073,1157,1165,1189,1241,1261,1285,1385,1417,1465,1565,1585,

%U 1649,1685,1765,1769,1781,1853,1865,1921,1937,1985,2117,2165,2173,2257,2285,2509,2561,2581,2785,2813,2885,2929,2941

%N Numbers of the form p*q where p, q are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = -1.

%C If k is a term, the Pell equation x^2 - k*y^2 = -1 has a solution [Dirichlet, Newman (1977)]. This is only a sufficient condition, there are many other solutions, see A031396.

%H Rémy Sigrist, <a href="/A322781/b322781.txt">Table of n, a(n) for n = 1..10000</a>

%H Morris Newman, <a href="https://www.jstor.org/stable/2319968">A note on an equation related to the Pell equation</a>, The American Mathematical Monthly 84.5 (1977): 365-366.

%o (PARI) isok(n) = my (f=factor(n)); omega(f)==2 && big omega(f)==2 && f[1,1]%4==1 && f[2,1]%4==1 && kronecker(f[1,1], f[2,1])==-1 \\ _Rémy Sigrist_, Jan 11 2019

%o (Python)

%o from sympy.ntheory import legendre_symbol, factorint

%o A322781_list, k = [], 1

%o while len(A322781_list) < 10000:

%o     fk, fv = zip(*list(factorint(4*k+1).items()))

%o     if sum(fv) == len(fk) == 2 and fk[0] % 4 == fk[1] % 4 == 1 and legendre_symbol(fk[0],fk[1]) == -1:

%o             A322781_list.append(4*k+1)

%o     k += 1 # _Chai Wah Wu_, Jan 11 2019

%Y Cf. A002144, A031396, A323271, A323272.

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Jan 11 2019