%I
%S 11,5,31,11,59,19,95,29,139,41,191,55,251,71,319,89,395,109,479,131,
%T 571,155,671,181,779,209,895,239,1019,271,1151,305,1291,341,1439,379,
%U 1595,419,1759,461,1931,505,2111,551,2299,599,2495,649,2699,701,2911,755
%N Sum of 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 Q = ax^2 + (4a/k  b)x + 4a/k^2 + 2b/k + c = 0. For a,b,c = 1,1,1 and k = 1,2,3..., the coefficients given by Q are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions and summing a+b+c gives the sequence.
%C The negative of the nth term is the n+4th term of the c coefficient sequence A229526.
%H Colin Barker, <a href="/A229525/b229525.txt">Table of n, a(n) for n = 1..1000</a>
%H Russell Walsmith, <a href="/A229526/a229526.pdf">CLChemy III: HyperQuadratics</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,3,0,1).
%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(n2)3*a(n4)+a(n6). G.f.: x*(x^5x^44*x^32*x^2+5*x+11) / ((x1)^3*(x+1)^3).  _Colin Barker_, Nov 02 2014
%F a(n) = (5+3*(1)^n)*(4+6*n+n^2)/8.  _Colin Barker_, Nov 03 2014
%e For k = 5, the coefficients are 1, 9/5, 11/25. Clearing fractions, 25, 45, 11 and 25 + 45 11 = 59 = a[5].
%o (PARI) Vec(x*(x^5x^44*x^32*x^2+5*x+11)/((x1)^3*(x+1)^3) + O(x^100)) \\ _Colin Barker_, Nov 02 2014
%Y The a coefficients are A168077, b coefficients are A171621, c coefficients are A229526.
%K nonn,easy
%O 1,1
%A _Russell Walsmith_, Sep 26 2013
%E More terms from _Colin Barker_, Nov 02 2014
