OFFSET
1,2
COMMENTS
The 2^n zeros of P(n,x) are 2+2*cos[(2k-1)Pi/(2^(n+1))], k=1,2,...,2^n.
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.
LINKS
Clark Kimberling, Polynomials defined by a second-order recurrence, interlacing zeros, and Gray codes, The Fibonacci Quarterly 48 (2010) 209-218.
FORMULA
P(n+1,x+2) = P(n,x^2) for n>=0.
EXAMPLE
Row 1: 1 -2 (from x-2)
Row 2: 1 -4 2 (from x^2-4x+2)
Row 3: 1 -8 20 -16 2
Row 4: 1 -16 104 -352 660 -672 336 -64 2
PROG
(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
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Clark Kimberling, Apr 02 2009
STATUS
approved