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%I #15 Nov 07 2014 15:44:48
%S 5,1,1,-1,-11,-5,-31,-11,-59,-19,-95,-29,-139,-41,-191,-55,-251,-71,
%T -319,-89,-395,-109,-479,-131,-571,-155,-671,-181,-779,-209,-895,-239,
%U -1019,-271,-1151,-305,-1291,-341,-1439,-379,-1595,-419,-1759,-461,-1931,-505
%N The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...
%C The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. Let a,b,c = 1,-1,-1 and k = 1,2,3... Then the coefficients given by this last equation are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions, the c coefficients are the sequence above.
%C The n-th term = the (positive) n-4th term of A229525.
%H Colin Barker, <a href="/A229526/b229526.txt">Table of n, a(n) for n = 1..1000</a>
%H Russell Walsmith, <a href="/A229526/a229526.pdf">CL-Chemy III: Hyper-Quadratics</a>
%F ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3..n.
%F a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5+x^4-4*x^3-14*x^2+x+5) / ((x-1)^3*(x+1)^3). - _Colin Barker_, Nov 02 2014
%F a(n) = (-5+3*(-1)^n)*(-4-2*n+n^2)/8. - _Colin Barker_, Nov 03 2014
%e For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions gives 25, 45, -11 and -11 = a[5].
%o (PARI) Vec(-x*(x^5+x^4-4*x^3-14*x^2+x+5)/((x-1)^3*(x+1)^3) + O(x^100)) \\ _Colin Barker_, Nov 02 2014
%Y The a coefficients are A168077, b coefficients are A171621, the sum of a, b and c coefficients is A229525.
%K sign,easy
%O 1,1
%A _Russell Walsmith_, Sep 27 2013
%E More terms from _Colin Barker_, Nov 02 2014
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