%I #12 Jan 28 2024 09:20:58
%S 0,2,11,207,99919,32416037103,4788545326929179011183,
%T 147201835861247697127798679336116306013028335,
%U 196331785117316517420778884783875086749917195699904294273499082962835791812062775501401839
%N Define P(n,X) by the recursion P(1,X) = 1, P(n+1,X) = (P(n,X)+X)^2; then a(1) = 0 and for n > 1 a(n) is the coefficient of X^(2^(n-2)) in P(n,X) of degree 2^(n-1).
%C a(n) is the greatest coefficient in P(n,X). Next term is too large to include.
%F For n > 4, 2^(2^(n-1)) < a(n) < (5/2)^(2^(n-1)).
%e P(1,X) = 1 then P(2,X) = (1+X)^2 = X^2+2X+1, the coefficient of X^(2^(2-2)) = X is 2 = a(2). P(4,X) = x^8+12*x^7+58*x^6+146*x^5+207*x^4+166*x^3+71*x^2+14*x+1 and the coefficient of X^(2^(4-2)) = X^4 is 207 = a(4).
%o (PARI) u=1; for(n=2,6,a=(u+x)^2; u=a; print1(polcoeff(u,2^(n-2),x),","))
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, May 09 2002