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A158985 Coefficients of polynomials (in descending powers of x) P(n,x) := 1 + P(n-1,x)^2, where P(1,x) = x + 1.

%I A158985 #19 Jan 05 2025 19:51:39
%S A158985 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 A158985 7664,18208,35296,56472,74944,82432,74624,54792,31776,13888,4160,677,
%U A158985 1,32,512,5440,43048,269920,1393728,6082752,22860480,75010560,217147904
%N A158985 Coefficients of polynomials (in descending powers of x) P(n,x) := 1 + P(n-1,x)^2, where P(1,x) = x + 1.
%H A158985 Clark Kimberling, <a href="https://web.archive.org/web/2024*/https://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 A158985 From _Peter Bala_, Jul 01 2015: (Start)
%F A158985 P(n,x) = P(n,-2 - x) for n >= 2.
%F A158985 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 A158985 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 A158985 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 A158985 n-th row sum = P(n,1) = A003095(n+1);
%F A158985 P(n,1) = P(n+1,0) = P(n+1,-2); P(n,1) = P(n,-3) for n >= 2.
%F A158985 P(n,2) = A062013(n). (End)
%e A158985 Row 1: 1 1 (from x + 1)
%e A158985 Row 2: 1 2 2 (from x^2 + 2*x + 2)
%e A158985 Row 3: 1 4 8 8 5
%e A158985 Row 4: 1 8 32 80 138 168 144 80 26
%o A158985 (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 A158985 Cf. A158982, A158983, A158984, A158986, A003095 (row sums), A062013.
%K A158985 nonn,tabf
%O A158985 1,4
%A A158985 _Clark Kimberling_, Apr 02 2009