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%I #27 Sep 08 2022 08:44:57
%S 2,3,7,10,11,15,18,19,23,26,27,31,34,35,39,42,43,47,50,51,55,58,59,63,
%T 66,67,71,74,75,79,82,83,87,90,91,95,98,99,103,106,107,111,114,115,
%U 119,122,123,127,130,131,135,138,139,143,146,147,151,154,155,159
%N Numbers that are congruent to {2, 3, 7} mod 8.
%C From _Richard R. Forberg_, Mar 14 2016 : (Start)
%C 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.
%C 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).
%C 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*k-3)^2 mod 48.
%C (End)
%H G. C. Greubel, <a href="/A047531/b047531.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).
%F From _Chai Wah Wu_, May 30 2016: (Start)
%F a(n) = a(n-1) + a(n-3) - a(n-4) for n > 4.
%F G.f.: x*(x^3 + 4*x^2 + x + 2)/(x^4 - x^3 - x + 1). (End)
%F 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
%F a(3k) = 8k-1, a(3k-1) = 8k-5, a(3k-2) = 8k-6. - _Wesley Ivan Hurt_, Jun 10 2016
%p A047531:=n->(24*n-12+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
%t LinearRecurrence[{1, 0, 1, -1}, {2, 3, 7, 10}, 50] (* _G. C. Greubel_, May 30 2016 *)
%t Select[Range[200],MemberQ[{2,3,7},Mod[#,8]]&] (* _Harvey P. Dale_, Jan 18 2019 *)
%o (Magma) [n : n in [0..150] | n mod 8 in [2, 3, 7]]; // _Wesley Ivan Hurt_, Jun 10 2016
%Y Cf. A047438.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_
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