

A047531


Numbers that are congruent to {2, 3, 7} mod 8.


1



2, 3, 7, 10, 11, 15, 18, 19, 23, 26, 27, 31, 34, 35, 39, 42, 43, 47, 50, 51, 55, 58, 59, 63, 66, 67, 71, 74, 75, 79, 82, 83, 87, 90, 91, 95, 98, 99, 103, 106, 107, 111, 114, 115, 119, 122, 123, 127, 130, 131, 135, 138, 139, 143, 146, 147, 151, 154, 155, 159
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OFFSET

1,1


COMMENTS

From Richard R. Forberg, Mar 14 2016 : (Start)
Numbers m such that p^2 + q^2 + r^2  m is a square, for distinct odd primes p, q, and r. This follows from the fact that the sum of the squares of three odd primes is 3 mod 8, and squares are {0,1,4} mod 8. Changing m to +m leads to A047438.
The expression p^2 + q^2 + r^2  m can equal the square of an odd number only when m == 2 mod 8 (because all odd squares are 1 mod 8).
This same expression only equals squares of primes for m in {2, 10, 26, 34} mod 48, because when m = {18, 42} mod 48, the expression is congruent to {9, 33} mod 48, which matches the values of (6*k3)^2 mod 48.
(End)


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).


FORMULA

From Chai Wah Wu, May 30 2016: (Start)
a(n) = a(n1) + a(n3)  a(n4) for n > 4.
G.f.: x*(x^3 + 4*x^2 + x + 2)/(x^4  x^3  x + 1). (End)
a(n) = (24*n + 5*sqrt(3)*sin(2*Pi*n/3) + 3*cos(2*Pi*n/3)  12)/9.  Ilya Gutkovskiy, May 30 2016
a(3k) = 8k1, a(3k1) = 8k5, a(3k2) = 8k6.  Wesley Ivan Hurt, Jun 10 2016


MAPLE

A047531:=n>(24*n12+3*cos(2*n*Pi/3)+5*sqrt(3)*sin(2*n*Pi/3))/9: seq(A047531(n), n=1..100); # Wesley Ivan Hurt, Jun 10 2016


MATHEMATICA

LinearRecurrence[{1, 0, 1, 1}, {2, 3, 7, 10}, 50] (* G. C. Greubel, May 30 2016 *)
Select[Range[200], MemberQ[{2, 3, 7}, Mod[#, 8]]&] (* Harvey P. Dale, Jan 18 2019 *)


PROG

(MAGMA) [n : n in [0..150]  n mod 8 in [2, 3, 7]]; // Wesley Ivan Hurt, Jun 10 2016


CROSSREFS

Cf. A047438.
Sequence in context: A051637 A051471 A140794 * A102808 A215937 A140512
Adjacent sequences: A047528 A047529 A047530 * A047532 A047533 A047534


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



