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%I #35 Nov 24 2024 01:51:27
%S 5,7,17,19,29,31,41,43,53,55,65,67,77,79,89,91,101,103,113,115,125,
%T 127,137,139,149,151,161,163,173,175,185,187,197,199,209,211,221,223,
%U 233,235,245,247,257,259,269,271,281,283,293,295,305,307,317,319,329,331
%N Numbers that are congruent to {5, 7} (mod 12).
%D L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 64.
%H Amiram Eldar, <a href="/A092242/b092242.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F 1/5^2 + 1/7^2 + 1/17^2 + 1/19^2 + 1/29^2 + 1/31^2 + ... = Pi^2*(2 - sqrt(3))/36 = 0.073459792... [Jolley] - _Gary W. Adamson_, Dec 20 2006
%F a(n) = 12*n - a(n-1) - 12 (with a(1)=5). - _Vincenzo Librandi_, Nov 16 2010
%F From _R. J. Mathar_, Oct 08 2011: (Start)
%F a(n) = 6*n - 3 - 2*(-1)^n.
%F G.f.: x*(5+2*x+5*x^2) / ( (1+x)*(x-1)^2 ). (End)
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (2 - sqrt(3))*Pi/12. - _Amiram Eldar_, Dec 04 2021
%F From _Amiram Eldar_, Nov 24 2024: (Start)
%F Product_{n>=1} (1 - (-1)^n/a(n)) = sec(Pi/12) (A120683).
%F Product_{n>=1} (1 + (-1)^n/a(n)) = (sqrt(3)/2)*sec(Pi/12) (= A010527 * A120683). (End)
%t Select[Range[331], MemberQ[{5, 7}, Mod[#, 12]] &] (* _Amiram Eldar_, Dec 04 2021 *)
%Y Fifth row of A092260.
%Y Cf. A010527, A120683.
%K nonn,easy
%O 1,1
%A _Giovanni Teofilatto_, Feb 19 2004
%E Edited and extended by _Ray Chandler_, Feb 21 2004