|
%I #18 Jan 17 2018 09:42:51
%S 1,1,1,0,3,1,-1,3,5,1,-1,-1,10,7,1,0,-6,7,21,9,1,1,-6,-10,31,36,11,1,
%T 1,1,-29,7,79,55,13,1,0,9,-24,-63,81,159,78,15,1,-1,9,15,-123,-54,264,
%U 279,105,17,1,-1,-1,57,-69,-321,132,624,447,136,19,1,0,-12,50,126,-459,-507,741,1245,671,171,21,1,1,-12,-20,302,-81,-1419,-258,2163,2227,959,210,23,1
%N Triangle read by rows of coefficients of polynomials Q_n(x) = 2^(-n)*((x + sqrt(x*(x + 6) - 3) + 1)^n - (x - sqrt(x*(x + 6) - 3) + 1)^n)/sqrt(x*(x + 6) - 3).
%C The polynomials Q_n(x) have generating function G(x,t) = t/(1 - (x + 1)*t - (x - 1)*t^2) = t + (x + 1)*t^2 + x*(x + 3)*t^3 + (x^3 + 5*x^2 + 3*x - 1)*t^4 + ...
%C Q_n(x) can be defined by the recurrence relation Q_n(x) = (x + 1)*Q_(n-1)(x) + (x - 1)*Q_(n-2)(x), Q_0(x)=0, Q_1(x)=1.
%C Discriminants of Q_n(x) gives the sequence: 0, 1, 1, 9, 320, 35600, 10948608, 8664190976, 16836271800320, 77757312009240576, 833309554769920000000, 20346889104219547132493824,...
%C Q_n(0) = A128834(n).
%C Q_n(1) = A000079(n-1), n>0.
%C Q_n(2) = A006190(n).
%C Q_n(3) = A090017(n).
%C Q_n(4) = A015536(n).
%C Q_n(5) = A135032(n).
%C Q_n(6) = A015562(n).
%C Q_n(7) = A190560(n).
%C Q_n(8) = A015583(n).
%C Q_n(9) = A190957(n).
%C Q_n(10) = A015603(n).
%H G. C. Greubel, <a href="/A271451/b271451.txt">Table of n, a(n) for the first 101 rows, flattened</a>
%H Ilya Gutkovskiy, <a href="/A271451/a271451.pdf">Polynomials Q_n(x)</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>
%e Triangle begins:
%e 1;
%e 1, 1;
%e 0, 3, 1;
%e -1, 3, 5, 1;
%e -1, -1, 10, 7, 1;
%e ...
%e The first few polynomials are:
%e Q_0(x) = 0;
%e Q_1(x) = 1;
%e Q_2(x) = x + 1;
%e Q_3(x) = x^2 + 3*x;
%e Q_4(x) = x^3 + 5*x^2 + 3*x - 1;
%e Q_5(x) = x^4 + 7*x^3 + 10*x^2 - x - 1,
%e ...
%t Flatten[Table[CoefficientList[((x + Sqrt[x (x + 6) - 3] + 1)^n - (x - Sqrt[x (x + 6) - 3] + 1)^n)/2^n/Sqrt[x (x + 6) - 3], x], {n, 0, 13}]]
%Y Cf. A049310.
%K sign,tabl,easy
%O 1,5
%A _Ilya Gutkovskiy_, Apr 08 2016
|
