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.

13, 17, 41, 53, 89, 97, 109, 149, 157, 229, 233, 257, 281, 313, 317, 337, 353, 373, 397, 401, 421, 433, 457, 461, 557, 569, 577, 601, 641, 709, 733, 769, 797, 809, 829, 853, 857, 881, 953, 997, 1013, 1021, 1049, 1061, 1097, 1153, 1193, 1201, 1213, 1229, 1277, 1297

Offset: 1

Author

Wolfdieter Lang, Dec 11 2016

Keywords

• nonn
• easy

Comments

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.

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.

Examples

			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.
		

Crossrefs

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

Formula

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).

Extensions

More terms from Jinyuan Wang, Apr 20 2025