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%I #31 Sep 08 2022 08:46:02
%S -1,0,0,2,3,6,2,12,15,20,6,30,35,42,12,56,63,72,20,90,99,110,30,132,
%T 143,156,42,182,195,210,56,240,255,272,72,306,323,342,90,380,399,420,
%U 110,462,483,506,132,552,575,600,156
%N a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.
%C The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
%C Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
%C c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
%C Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
%C e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
%C f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
%C f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
%C e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
%C Then c(n) = e(n)*f(n).
%C Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
%C After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
%C Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
%C A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
%C Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
%C g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
%C h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
%C h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
%C k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
%C l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
%C k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
%C 2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
%C Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
%C For a(n), we divide by 4.
%C Note that A214297 is the reduced numerator of 1/4 - 1/A061038(n).
%C Note also that A168077(n) = A026741(n)^2.
%H Colin Barker, <a href="/A214630/b214630.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
%F a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
%F a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
%F a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
%F a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
%F a(n) = g(n) * h(n).
%F a(n) = k(n) * l(n).
%F G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - _Colin Barker_, Jan 22 2015
%F From _Luce ETIENNE_, Apr 08 2017: (Start)
%F a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
%F a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)
%t CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* _G. C. Greubel_, Sep 20 2018 *)
%t LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* _Harvey P. Dale_, Jul 01 2019 *)
%o (PARI) Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ _Colin Barker_, Jan 22 2015
%o (Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // _G. C. Greubel_, Sep 20 2018
%Y Cf. A000466, A002378, A131723.
%K sign,frac,easy
%O 0,4
%A _Paul Curtz_, Jul 23 2012
%E Edited by _N. J. A. Sloane_, Aug 04 2012
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