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

1, 1, 1, 2, 2, 1, 4, 8, 8, 5, 1, 8, 32, 80, 138, 168, 144, 80, 26, 1, 16, 128, 672, 2580, 7664, 18208, 35296, 56472, 74944, 82432, 74624, 54792, 31776, 13888, 4160, 677, 1, 32, 512, 5440, 43048, 269920, 1393728, 6082752, 22860480, 75010560, 217147904

Offset

1,4

Links

Table of n, a(n) for n=1..47.

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,(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).

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.

n-th row sum = P(n,1) = A003095(n+1);

P(n,1) = P(n+1,0) = P(n+1,-2); P(n,1) = P(n,-3) for n >= 2.

P(n,2) = A062013(n). (End)

Example

Row 1: 1 1 (from x + 1)
Row 2: 1 2 2 (from x^2 + 2*x + 2)
Row 3: 1 4 8 8 5
Row 4: 1 8 32 80 138 168 144 80 26

Prog

(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

Crossrefs

Cf. A158982, A158983, A158984, A158986, A003095 (row sums), A062013.

Sequence in context: A084606 A304209 A137399 * A295687 A087854 A185041

Adjacent sequences: A158982 A158983 A158984 * A158986 A158987 A158988

Keyword

nonn,tabf

Author

Clark Kimberling, Apr 02 2009

Status

approved