login
Coefficients of polynomials P(n,x):=-2+P(n-1,x)^2, where P(0,x)=x-2.
4

%I #20 Jan 05 2025 19:51:39

%S 1,-2,1,-4,2,1,-8,20,-16,2,1,-16,104,-352,660,-672,336,-64,2,1,-32,

%T 464,-4032,23400,-95680,283360,-615296,980628,-1136960,940576,-537472,

%U 201552,-45696,5440,-256,2,1,-64,1952,-37760,520144,-5430656,44662464

%N Coefficients of polynomials P(n,x):=-2+P(n-1,x)^2, where P(0,x)=x-2.

%C The 2^n zeros of P(n,x) are 2+2*cos[(2k-1)Pi/(2^(n+1))], k=1,2,...,2^n.

%C P(n,x) = 2*T(2^(n+1),(1/2)x^(1/2)), where T(k,t) is the k-th Chebyshev polynomial of the first kind.

%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 P(n+1,x+2) = P(n,x^2) for n>=0.

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

%e Row 2: 1 -4 2 (from x^2-4x+2)

%e Row 3: 1 -8 20 -16 2

%e Row 4: 1 -16 104 -352 660 -672 336 -64 2

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

%Y Cf. A084534, A158983, A158984, A158985, A158986.

%K sign,tabf

%O 1,2

%A _Clark Kimberling_, Apr 02 2009