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%I #13 Mar 01 2016 03:25:37
%S 1,2,1,4,6,1,8,28,48,38,1,16,120,544,1628,3296,4432,3648,1446,1,32,
%T 496,4928,35064,189248,800992,2711424,7419740,16475584,29610272,
%U 42666880,48398416,41867904,26125248,10550016,2090918,1,64,2016,41600,631536
%N Coefficients of polynomials (in descending powers of x) P(n,x) := 2 + P(n-1,x)^2, where P(1,x) = x + 2.
%H Clark Kimberling, <a href="http://www.fq.math.ca/Papers1/48-3/Kimberling.pdf">Polynomials defined by a second-order recurrence, interlacing zeros, and Gray codes</a>, The Fibonacci Quarterly 48 (2010) 209-218.
%F From _Peter Bala_, Jul 01 2015: (Start)
%F P(n,x) = P(n,-4 - x) for n >= 2.
%F P(n+1,x)= P(n,(2 + x)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) - 2 is a zero of P(n+1,x).
%F Define a sequence of polynomials Q(n,x) by setting Q(1,x) = 2 + x^2 and Q(n,x) = Q(n-1, 2 + x^2) for n >= 2. Then P(n,x) = Q(n,sqrt(x)).
%F Q(n,x) = Q(k,Q(n-k,x)) for 1 <= k <= n-1; P(n,x) = P(k,P(n-k,x)^2) for 1 <= k <= n - 1.
%F n-th row sum = P(n,1) = A102847(n);
%F P(n,1) = P(n+1,-1) = P(n+1,-3); P(n,1) = P(n,-5) for n >= 2.
%F (End)
%e Row 1: 1 2 (from x+2)
%e Row 2: 1 4 6 (from x^2+4x+6)
%e Row 3: 1 8 28 48 38
%e Row 4: 1 16 120 544 1628 3296 4432 3648 1446
%Y Cf. A158982, A158984, A158985, A158986. A102847 (row sums).
%K nonn,tabf
%O 1,2
%A _Clark Kimberling_, Apr 02 2009
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