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%I #17 Mar 01 2016 03:25:28
%S 1,1,1,2,2,1,4,8,8,5,1,8,32,80,138,168,144,80,26,1,16,128,672,2580,
%T 7664,18208,35296,56472,74944,82432,74624,54792,31776,13888,4160,677,
%U 1,32,512,5440,43048,269920,1393728,6082752,22860480,75010560,217147904
%N Coefficients of polynomials (in descending powers of x) P(n,x) := 1 + P(n-1,x)^2, where P(1,x) = x + 1.
%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,-2 - x) for n >= 2.
%F P(n+1,x)= P(n,(1 + x)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) - 1 is a zero of P(n+1,x).
%F Define a sequence of polynomials Q(n,x) by setting Q(1,x) = 1 + x^2 and Q(n,x) = Q(n-1, 1 + 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) = A003095(n+1);
%F P(n,1) = P(n+1,0) = P(n+1,-2); P(n,1) = P(n,-3) for n >= 2.
%F P(n,2) = A062013(n). (End)
%e Row 1: 1 1 (from x + 1)
%e Row 2: 1 2 2 (from x^2 + 2*x + 2)
%e Row 3: 1 4 8 8 5
%e Row 4: 1 8 32 80 138 168 144 80 26
%o (PARI) tabf(nn) = {my(P = x+1); print(Vec(P)); for (n=1, nn, P = 1 + P^2; print(Vec(P)););} \\ _Michel Marcus_, Jul 01 2015
%Y Cf. A158982, A158983, A158984, A158986, A003095 (row sums), A062013.
%K nonn,tabf
%O 1,4
%A _Clark Kimberling_, Apr 02 2009
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