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%I #45 Jun 29 2021 10:35:35
%S 0,0,1,3,10,15,45,120,351,528,1540,4095,11935,17955,52326,139128,
%T 405450,609960,1777555,4726275,13773376,20720703,60384555,160554240,
%U 467889345,703893960,2051297326,5454117903,15894464365,23911673955,69683724540,185279454480
%N Sequences A124174 and A006454 interlaced.
%C a(2n) and 2*a(2n) + 1 are triangular.
%C a(2n + 1) is triangular and a(2n + 1)/2 is the harmonic mean of consecutive triangular numbers (therefore, a(2n + 1) + 1 is square).
%H Colin Barker, <a href="/A217278/b217278.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,34,0,-34,0,-1,0,1).
%F a(n) = 35*(a(n-4) - a(n-8)) + a(n-12).
%F lim n --> infinity a(2n)/a(2n - 1) = (3 + sqrt(8))/2.
%F From _Raphie Frank_, Dec 21 2015: (Start)
%F a(2*n + 1) = 1/64*(((4 + sqrt(2)) * (1 - (-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2)) + (4 - sqrt(2)) * (1+(-1)^(n+1) * sqrt(2))^(2*floor((n+1)/2))))^2 - 1.
%F a(2*n + 2) = 1/2*(3*(a(2*n + 1)) + sqrt((a(2*n + 1)) + 1) * sqrt(8*(a(2*n + 1)) + 1) + 1).
%F (End)
%e a(18) = 35*(52326 - 1540) + 45 = 1777555,
%e a(19) = 35*(139128 - 4095) + 120 = 4726275.
%t LinearRecurrence[{0,1,0,34,0,-34,0,-1,0,1},{0,0,1,3,10,15,45,120,351,528},40] (* _Harvey P. Dale_, Aug 04 2019 *)
%o (PARI) concat([0,0], Vec(-x^2*(3*x^5+x^4+12*x^3+9*x^2+3*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)) + O(x^100))) \\ _Colin Barker_, Jun 23 2015
%Y Cf. (sqrt(8a(2n) + 1) - 1)/2 = A216134(n) = A216162(2n + 1).
%Y Cf. sqrt(a(2n+1) + 1) = A006452(n + 1) = A216162(2n + 2).
%Y Cf. (sqrt(8a(2n+1) + 1) - 1)/2 = A006451(n).
%K nonn,easy
%O 0,4
%A _Raphie Frank_, Sep 29 2012
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