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A158984 Coefficients of polynomials (in descending powers of x) P(n,x) := -1 + P(n-1,x)^2, where P(1,x) = x - 1. 4
1, -1, 1, -2, 0, 1, -4, 4, -1, 1, -8, 24, -32, 14, 8, -8, 0, 0, 1, -16, 112, -448, 1116, -1744, 1552, -384, -700, 736, -160, -128, 64, 0, 0, 0, -1, 1, -32, 480, -4480, 29112, -139552, 509600, -1441024, 3166616, -5345344, 6668992, -5473536, 1494624 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

LINKS

Table of n, a(n) for n=1..48.

Clark Kimberling, Polynomials defined by a second-order recurrence, interlacing zeros, and Gray codes, The Fibonacci Quarterly 48 (2010) 209-218.

FORMULA

From Peter Bala, Jul 01 2015: (Start)

P(n,x) = P(n,2 - x) for n >= 2.

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).

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)).

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.

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

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

EXAMPLE

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

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

Row 3: 1 -4 4 -1

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

PROG

(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

CROSSREFS

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

Sequence in context: A098689 A288515 A264583 * A158417 A139435 A077909

Adjacent sequences:  A158981 A158982 A158983 * A158985 A158986 A158987

KEYWORD

sign,tabf

AUTHOR

Clark Kimberling, Apr 02 2009

STATUS

approved

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Last modified July 31 13:25 EDT 2021. Contains 346373 sequences. (Running on oeis4.)