

A057127


2 is a square mod n.


12



1, 2, 3, 6, 9, 11, 17, 18, 19, 22, 27, 33, 34, 38, 41, 43, 51, 54, 57, 59, 66, 67, 73, 81, 82, 83, 86, 89, 97, 99, 102, 107, 113, 114, 118, 121, 123, 129, 131, 134, 137, 139, 146, 153, 162, 163, 166, 171, 177, 178, 179, 187, 193, 194, 198, 201, 209, 211, 214, 219
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OFFSET

1,2


COMMENTS

Includes the primes in A033203 and these (primes congruent to {1, 2, 3} mod 8) are the prime factors of the terms in this sequence.
Numbers that are not multiples of 4 and for which all odd prime factors are congruent to {1, 3} mod 8.  Eric M. Schmidt, Apr 21 2013
Positive integers primitively represented by x^2 + 2y^2.  Ray Chandler, Jul 22 2014
The set of the divisors of numbers of the form k^2+2.  Michel Lagneau, Jun 28 2015
The number of proper solutions (x, y) with nonnegative x of the positive definite primitive quadratic form x^2 + 2*y*2 (discriminant 8) representing a(n) is 1 for n = 1 and for n >= 2 it is 2^(P_1 + P_3), where P_1 and P_3 are the number of distinct prime divisors of a(n) congruent to 1 and 3 modulo 8, respectively. See the above comments on A033203 and this binary form.  Wolfdieter Lang, Feb 25 2021


LINKS

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)


EXAMPLE

Binary quadratic form x^2 + 2*y^2 representing a(n), with x >= 0: a(1) = 1: one solution (x, y) = (1,0); a(2) = 2: one solution (0,1); a(3) = 3: two solutions (1, pm 1), with pm = +1 or 1; a(5) = 9 = 3^2: two solutions (1, pm 2); a(12) = 33 = 3*11: 4 solutions (1, pm 4) and (5, pm 2); a(137) = 3*11*17 = 561: eight solutions (7, pm 16), (13, pm 14), (19, pm 10) and (23, pm 4).  Wolfdieter Lang, Feb 25 2021


MAPLE

select(n > numtheory:msqrt(2, n) <> FAIL, [$1..1000]); # Robert Israel, Jun 29 2015


MATHEMATICA

Select[Range[300], IntegerQ[PowerMod[2, 1/2, #]]&] // Quiet (* JeanFrançois Alcover, Mar 04 2019 *)


PROG

(Sage)
def isA057127(n):
if n % 4 == 0: return False
return all(p % 8 in [1, 2, 3] for p, _ in factor(n))
[n for n in range(1, 300) if isA057127(n)]
# Eric M. Schmidt, Apr 21 2013
(PARI) isok(n) = issquare(Mod(2, n)); \\ Michel Marcus, Jun 28 2015


CROSSREFS

Cf. A008784, A033203, A057125, A057126, A057128, A057129.
Sequence in context: A227000 A338546 A176189 * A224449 A007780 A018537
Adjacent sequences: A057124 A057125 A057126 * A057128 A057129 A057130


KEYWORD

nonn


AUTHOR

Henry Bottomley, Aug 10 2000


STATUS

approved



