A109086 Recursively defined polynomials, read by row.

1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 1, 0, 4, 2, 7, 4, 6, 2, 1, 1, 0, 8, 4, 30, 25, 72, 68, 116, 102, 118, 84, 67, 33, 16, 4, 1, 1, 0, 16, 8, 124, 114, 641, 776, 2495, 3372, 7637, 10444, 18561, 24212, 35684, 42828, 53707, 57884, 62257, 59056, 54261, 44356, 34326, 23548

Offset

1,8

Comments

lim(N->infinity)p_N(x)/prod(n = 0, N-1)p_n(x) = f(x), with f(x) from A109087.

Links

Robert Israel, Table of n, a(n) for n = 1..9957

Formula

p_0(x) = x, p_(n+1)(x) = p_n(x)^2 + prod(i = 0, n-1)p_i(x)^(n-i), n >= 0. p_n(x) = sum(i = 0, 2^n)t(n, i)*x^i, n >= 0. a(2^(n+1)+n-i)=t(n, i), n>=0, 0<=i<=2^n.

Example

p_0(x) = x, p_1(x) = x^2 + 1, p_2(x) = x^4 + 2*x^2 + x + 1,
p_3(x) = x^8 + 4*x^6 + 2*x^5 + 7*x^4 + 4*x^3 + 6*x^2 + 2*x + 1
p_4(x) = x^16 + 8*x^14 + 4*x^13 + 30*x^12 + 25*x^11 + 72*x^10 + 68*x^9 + 116*x^8 + 102*x^7 + 118*x^6 + 84*x^5 + 67*x^4 + 33*x^3 + 16*x^2 + 4*x + 1

Maple

p[0]:= x:
for n from 1 to 7 do
  p[n]:= expand(p[n-1]^2 + mul(p[i]^(n-1-i), i=0..n-2))
od:
for n from 0 to 7 do
  seq(coeff(p[n], x, j), j=degree(p[n]) .. 0, -1)
od; # _Robert Israel_, Jun 23 2020

Prog

(PARI) N=5; p=x; q=1; r=1; for(n=0, N, print(Vec(p)); q*=r; r*=p; p=p^2+q)

Crossrefs

Cf. A109088, A109089, A109090.

Sequence in context: A111941 A153462 A126310 * A213620 A331734 A105794

Adjacent sequences: A109083 A109084 A109085 * A109087 A109088 A109089

Keyword

easy,nonn,tabf,look

Author

Nikolaus Meyberg (Nikolaus.Meyberg(AT)t-online.de), Jun 19 2005

Status

approved