%I
%S 1,1,2,4,7,11,15,20,26,32,39,47,55,64,74,84,95,107,119,132,146,160,
%T 175,191,207,224,242,260,279,299,319,340,362,384,407,431,455,480,506,
%U 532,559,587,615,644,674,704,735,767,799,832,866,900,935,971,1007,1044
%N Number of distinct values taken by 3rd derivative of x^x^...^x (with n x's and parentheses inserted in all possible ways) at x=1.
%C Number of distinct values taken by 0th and 1st derivative is 1,1,1,1,1,1,... and by 2nd is 1,1,2,3,4,5,...
%F Conjectured g.f.: (x^62*x^4x^2+x1)*x/(x^52*x^4+x^3x^2+2*x1).  _Alois P. Heinz_, Nov 02 2011
%e For n=5 there are 7 distinct values: 9, 15, 18, 21, 24, 33, 48, they are given by 3rd derivatives of the following parenthesizations at x=1: x^(x^((x^x)^x)), x^((x^(x^x))^x), (x^x)^(x^(x^x)), x^(((x^x)^x)^x), (x^(x^x))^(x^x), (((x^(x^x)))^x)^x, (((x^x)^x)^x)^x. So a(5)=7.
%p f:= proc(n) option remember;
%p `if`(n=1, {[0, 0]}, {seq(seq(seq([2+g[1], 3*(1+g[1]+h[1])+g[2]],
%p h=f(nj)), g=f(j)), j=1..n1)})
%p end:
%p a:= n> nops(map(x> x[2], f(n))):
%p seq(a(n), n=1..40); # _Alois P. Heinz_, Nov 03 2011
%t f[1] = {x}; f[n_] := Flatten[Table[Outer[Power, f[k], f[nk]], {k, n1}]]; Table[Length[Union[D[f[n], {x, 3}] /. x > 1]], {n, 1, 8}] (* Reshetnikov *)
%t Table[If[n<3,1,Floor[(n^22)/3]],{n,1,100}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 01 2012 *)
%Y Cf. A000081 (upper bound), A000108, A199205 (4th derivatives), A199296 (5th derivatives), A215703, A215842. Column k=3 of A216368.
%K nonn
%O 1,3
%A _Vladimir Reshetnikov_, Nov 02 2011
%E a(13)a(56) from _Alois P. Heinz_, Nov 02 2011
