A158984 - OEIS

%I #18 Jan 05 2025 19:51:39

%S 1,-1,1,-2,0,1,-4,4,-1,1,-8,24,-32,14,8,-8,0,0,1,-16,112,-448,1116,

%T -1744,1552,-384,-700,736,-160,-128,64,0,0,0,-1,1,-32,480,-4480,29112,

%U -139552,509600,-1441024,3166616,-5345344,6668992,-5473536,1494624

%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="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 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,(x - 1)^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 P(n,x)^(2^k) divides P(n + 2*k,x) in Z[x] for k = 1,2,....

%F P(n,4) = A003096(n). (End)

%e Row 1: 1 -1 (from x-1)

%e Row 2: 1 -2 0 (from x^2-2x)

%e Row 3: 1 -4 4 -1

%e Row 4: 1 -8 24 -32 14 8 -8 0 0

%o (PARI) tabf(nn) = {p = x-1; print(Vec(p)); for (n=2, nn, p = -1 + p^2; print(Vec(p)););} \\ _Michel Marcus_, Mar 01 2016

%Y Cf. A158982, A158983, A158985, A158986, A003096.

%K sign,tabf

%O 1,4

%A _Clark Kimberling_, Apr 02 2009