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This is the union of {3}, A003627 (primes congruent to 2 mod 3) and A014752 (primes of the form x^2+27y^2). By Thm. 4.15 of [Cox], p is of the form x^2+27y^2 if and only if p is congruent to 1 mod 3 and 2 is a cubic residue mod p. If p is not congruent to 1 mod 3, then every number is a cubic residue mod p, including 2. - Andrew V. Sutherland, Apr 26 2008

Complement of A040034 relative to A000040. - Vincenzo Librandi, Sep 13 2012

Complement of A038873 relative to A000040.

Also primes p such that p divides 2^((p-1)/2) + 1. - Cino Hilliard, Sep 04 2004

Primes p such that p^2 == 25 (mod 48), n > 1. - Gary Detlefs, Dec 29 2011

This sequence gives the primes p which satisfy C(p, x = 0) = -1, where C(p, x) is the minimal polynomial of 2*cos(Pi/p) (see A187360). For a proof see a comment on C(n, 0) in A230075. - Wolfdieter Lang, Oct 24 2013

Except for the initial 3, these are the primes p such that Fibonacci(p) mod 6 = 5. - Gary Detlefs, May 26 2014

Inert rational primes in the field Q(sqrt(2)). - N. J. A. Sloane, Dec 26 2017

If a prime p is congruent to 3 or 5 (mod 8) and r > 1, then 2^((p-1)*p^(r-1)/2) == -1 (mod p^r). - Marina Ibrishimova, Sep 29 2018

For the proofs or the comments by Cino Hilliard and Marina Ibrishimova, see link below. - Robert Israel, Apr 24 2019

For a prime p congruent to 1 mod 8, 2 is a biquadratic residue mod p if and only if there are integers x,y such that x^2 + 64*y^2 = p. 2 is also a biquadratic residue mod 2 and mod p for any prime p congruent to 7 mod 8 and for no other primes. - Fred W. Helenius (fredh(AT)ix.netcom.com), Dec 30 2004

Complement of A040100 relative to A000040. - Vincenzo Librandi, Sep 13 2012

Or, primes p which do not divide 2^n+1 for any n.

The possibility n=0 in the above rules out A072936(1)=2; apart from this, a(n)=A072936(n+1). - M. F. Hasler, Dec 08 2007

The order of 2 mod p is odd iff 2^k=1 mod p, where p-1=2^s*k, k odd. - M. F. Hasler, Dec 08 2007

Has density 7/24 (Hasse).

From Jianing Song, Jun 27 2025: (Start)

The multiplicative order of 2 modulo a(n) is A139686(n).

Contained in primes congruent to 1 or 7 modulo 8 (primes p such that 2 is a quadratic residue modulo p, A001132), and contains primes congruent to 7 modulo 8 (A007522). (End)

For the fundamental positive solution x(n)^2 - a(n)*y(n)^2 = 2 see (x(n) = A261247(n), y(n) = A261248(n)), for n >= 1.

Conjecture: The sequence consists of all numbers D not a square and even D = 2*d has odd d with prime factors of the form 1 or 7 (mod 8). Odd D has prime factors of the form 1 or 7 (mod 8) but there is an odd number of primes of the form 7 (mod 8). The following will prove that these conditions for D are necessary in order to have solutions.

This conjecture is false. For the odd D case see the counterexamples in A263010, and for the even D in A264352. - Wolfdieter Lang, Nov 12 2015

If there is a solution for D, D not a square, then only one class of solution exists due to Nagell's Theorem 110, p. 208, because then 2 divides 2*D. All solutions will be proper because 2 is a prime.

For the even prime D = p = 2 the positive fundamental solution is [x(1) = 2, y(1) = 1].

For odd primes D = p there can be solutions only for p == +7 (mod 8), that is p from A007522. Then x and y are both odd. Proof: Consider a solution of x^2 - p*y^2 = 2. The parities of x and y have to be either even and even or odd and odd. For odd x one has x^2 == +1 (mod 8) (because x^2 = 8*T(X) + 1 with x = 2*X+1 and the triangular numbers T = A000217); similarly for y^2 if y is odd. In the even-even case x^2 and y^2 are both congruent to 4 (mod 8). The even-even case leads to 4 - 4*p = 2 (mod 8), excluding all odd p, namely p == 1, 3, 5, 7 (mod 8). The odd-odd case is 1 - p*1 = 2 (mod 8), and p == 1, 3, 5 (mod 8) are excluded. Therefore, only p == 7 (mod 8) qualifies for a solution, and then x and y will be both odd.

For D = p == 7 (mod 8) from A007522 one can test if there exists a fundamental positive solution (at most one class can exist, therefore there is either no solution or just one) [2*U(p)+1, 2*V(p)+1] by checking the two inequalities (see Nagell, eq. (4) and (5), p. 206) 0 <= V(p) < floor((Y(p)/sqrt(X(p) + 1) - 1)/2) and 0 <= U(p) <= floor((sqrt(X(p) + 1) - 1)/2), with the positive fundamental solution [X(p), Y(p)] of X^2 - p*Y^2 = +1. These solutions can be found in (A033313(k), A033317(k)) if A000037(k) is the prime p == 7 (mod 8) one is testing.

For composite even D there are solutions only if D/2 is odd. Proof: If D is even then x has to be even, hence x^2 == 0 (mod 4) and then D*y^2 == -2 (mod 4), hence D cannot be 0 (mod 4). Thus an even D can only be of the form D = 2*d with d odd. The modulo 3 and modulo 5 argument used in the next case will show that d can have only prime factors of the form +1 or -1 (mod 8).

For composite odd D one finds like above that the even-even x and y case is excluded, and the odd-odd case needs D == -1 (mod 8) == 7 (mod 8). Hence a candidate for D is from A004771 - A007522. D cannot have any prime factor p of the form 3 or 5 (mod 8) because otherwise x^2 == 2 (mod p), but the Legendre symbol (2/p) = -1 for such p's (see, e.g., Nagell, Theorem 81, p. 136). For example, D = 15 = 3*5 cannot have a solution. Thus the only candidates for D have prime factors p of the form +1 or +7 (mod 8), with the number of the latter ones being odd. E.g., D = 7*17 = 119 qualifies as a candidate and it has indeed solutions, namely the ones obtainable from the fundamental one [11, 1].

The general proper positive solutions for D(n) = a(n) are obtained from the fundamental ones [x(n), y(n)] given in A261247 and A261248 with the help of powers of the matrix M(n) = [[u(n), D(n)*v(n)], [v(n), u(n)]], where u(n) and v(n) are the positive fundamental solutions of U(n) - D(n)*V(n) = 1, by (x(n; k), y(n; k))^T = M(n)^k (x(n), y(n))^T (T for transposed), for k >= 0. [u(n), v(n)] = [A033313(j(n)), A033317(j(n))] if A000037(j(n)) = D(n) = a(n).

Observation: All degrees (7, 47, 79, 103, 119, 127) of the modular equations derived for solving Ramanujan's question 699 by Galkin & Kozirev (see reference and A318732) are terms of this sequence. - Hugo Pfoertner, Sep 24 2023

Coincides with the sequence of "primes p such that x^16 = 2 has a solution mod p" for first 58 terms (and then diverges).

Complement of A045316 relative to A000040. - Vincenzo Librandi, Sep 13 2012

The prime numbers congruent to +1 or -1 modulo 8 of this sequence appear exactly once. For a proof see the W. Lang link under A001132. - Wolfdieter Lang, Feb 17 2015

The prime numbers in this sequence define A001132 (see comment in A001132). - Richard Choulet, Dec 16 2008

For the sum of legs of Pythagorean triangles with multiple entries see A198390. - Wolfdieter Lang, May 24 2013

Are these just the positive multiples of A001132? - Charles R Greathouse IV, May 28 2013

For the sum of legs of primitive Pythagorean triangles see A120681. - Wolfdieter Lang, Feb 17 2015

n is in the sequence iff A331671(n) > 0. - Ray Chandler, Feb 26 2020

There is a connection to the leg sums of Pythagorean triangles.

See a comment on the primitive case under A198439, which applies mutatis mutandis. - Wolfdieter Lang, May 23 2013

Are these just the positive multiples of A001132? - Charles R Greathouse IV, May 28 2013

n appears A331671(n) times. - Ray Chandler, Feb 26 2020