%I
%S 0,1,1,1,2,3,6,12,24,50,107,232,508,1124,2513,5665,12858,29356,67371,
%T 155345,359733,836261,1950829,4565305,10714501,25212843,59474318,
%U 140609809,333126672,790764280,1880489541,4479494059,10687448937,25536624382,61102431113
%N Define polynomials Pn by P0 = 0, P1 = x, P2 = P1 + P0, P3 = P2 * P1, P4 = P3 + P2, etc. alternately adding or multiplying. For even n > 2k, then first k coefficients of Pn remain unchanged and their values are the first k terms of the sequence.
%H Alois P. Heinz and Vaclav Kotesovec, <a href="/A045761/b045761.txt">Table of n, a(n) for n = 0..2000</a> (terms 0..1000 from Alois P. Heinz)
%H JeanLuc Baril, Sergey Kirgizov, Armen Petrossian, <a href="http://jl.baril.ubourgogne.fr/decreasing.pdf">Dyck paths with a first return decomposition constrained by height</a>, Submitted, 2017.
%F a(n) ~ c * d^n / n^(3/2), where d = 2.50297436517909273228379630... and c = 0.34042564735836570861482... .  _Vaclav Kotesovec_, Aug 08 2016, updated Aug 27 2016
%F Conjecture: 1/d = 0.39952466709679946... = A268107.  _JeanFrançois Alcover_, Aug 08 2016
%e The sequence of polynomials is 0, x, x, x^2, x^2 + x, x^4 + x^3, x^4 + x^3 + x^2 + x, ..., and after this all the even polynomials end with x^3 + x^2 + x (+ 0), so the first 4 terms of the sequence are these coefficients (in ascending order): 0, 1, 1, 1.  _Michael B. Porter_, Aug 09 2016
%t k = 32; P[0] = 0; P[1] = x;
%t P[n_] := P[n] = If[EvenQ[n], P[n1] + P[n2], P[n1]*P[n2]] + O[x]^(2k+1) // Normal;
%t CoefficientList[P[2k], x][[1 ;; k+1]] (* _JeanFrançois Alcover_, Aug 07 2016 *)
%Y Cf. A001696, A039941, A268107.
%K nonn
%O 0,5
%A James Boudinot (jboudinot(AT)yahoo.com)
%E More terms from _Michael Somos_, May 19 2000
