A279392 - OEIS

%I #10 May 27 2024 07:16:42

%S 13,17,41,53,89,97,109,149,157,229,233,257,281,313,317,337,353,373,

%T 397,401,421,433,457,461,557,569,577,601,641,709,733,769,797,809

%N Bisection of primes congruent to 1 modulo 4 (A002144), depending on the corresponding sum of the A002972 and 2*A002973 entries being congruent to 1 modulo 4 or not. Here we give the first case.

%C The primes from A002144 (1 (mod 4) primes) have the property A002144(n) = A002972(n)^2 + (2*A002973(n))^2 = A(n)^2 + B(n)^2 with odd A(n) and even B(n). A bisection of A002144 is given depending on A(n) + B(n) == 1 (mod 4) (part I) or A(n) + B(n) == 3 (mod 4) (part II). The present sequence gives part I of this bisection. The other part II is given in A279393.

%C This bisection appears in the formula for the p-defects of the congruence y^2 == x^3 + 4*x (mod p) for primes p == 1 (mod 4). See A278720 where for nonvanishing entries the sign is conjectured to be + for these part I primes, and it is - for the part II primes from A279393.

%F A prime A002144(m) = A(m)^2 + B(m)^2 belongs to this sequence iff (-1)^((A(m)-1)/2 + B(m)/2) = +1, where A(m) = A002972(m) and B(m)/2 = A002973(m).

%e a(1) = 13 is the first prime from A002144 which has A + B = 1 (mod 4) because 13 = A002144(2) = A(2)^2 + B(2)^2 = 3^2 + (2*1)^2, and 3 + 2 = 5 == 1 (mod 4), and A002144(1) = 5 leads to  A + B = 3 (mod 4), because 5  = 1^2 + (2*1)^2.

%Y Cf. A279393, A002144, A002972, A002973, A278720.

%K nonn,easy,more

%O 1,1

%A _Wolfdieter Lang_, Dec 11 2016