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A279392 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.

%I A279392 #14 Apr 20 2025 20:13:45
%S A279392 13,17,41,53,89,97,109,149,157,229,233,257,281,313,317,337,353,373,
%T A279392 397,401,421,433,457,461,557,569,577,601,641,709,733,769,797,809,829,
%U A279392 853,857,881,953,997,1013,1021,1049,1061,1097,1153,1193,1201,1213,1229,1277,1297
%N A279392 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 A279392 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 A279392 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 A279392 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 A279392 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 A279392 Cf. A002144, A002972, A002973, A278720, A279393.
%K A279392 nonn,easy
%O A279392 1,1
%A A279392 _Wolfdieter Lang_, Dec 11 2016
%E A279392 More terms from _Jinyuan Wang_, Apr 20 2025