A216368 Number T(n,k) of distinct values taken by k-th derivative of x^x^...^x (with n x's and parentheses inserted in all possible ways) at x=1; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 7, 9, 9, 1, 5, 11, 17, 20, 20, 1, 6, 15, 30, 45, 48, 48, 1, 7, 20, 50, 92, 113, 115, 115, 1, 8, 26, 77, 182, 262, 283, 286, 286, 1, 9, 32, 113, 342, 591, 691, 717, 719, 719, 1, 10, 39, 156, 601, 1263, 1681, 1815, 1838, 1842, 1842

Offset: 1

Author

Alois P. Heinz, Sep 05 2012

Keywords

• nonn
• tabl

Comments

T(n,k) <= A000081(n) because there are only A000081(n) different functions that can be represented with n x's.

It is not true that T(n,n) = T(n,n-1) for all n>1: T(13,13) - T(13,12) = 12486 - 12485 = 1.

Conjecture: T(n,n) = A000081(n) for n>=1. It would be nice to have a proof (or a disproof if the conjecture is wrong).

From Bradley Klee, Jun 01 2015 (Start):

I made a descendant graph (Plot 1) that shows how each derivative relates to the next. In this picture the number of nodes in row k gives the value T(n,k). You can see at n=6 collisions begin to occur, and at n=7 the situation is even worse. I then computed a new triangle with collisions removed (Plot 2) and values:

1

1 1

1 2 2

1 3 4 4

1 4 7 9 9

1 5 11 88 20 20

1 6 16 34 46 48 48

I suspect that Plot 2 will admit a recursive construction more readily than the graphs with collisions. You can already see that each graph "n-1" is a subgraph of graph "n" and that the remainder of graph "n" is similar to graph "n-1" with additional branches. (End)

Examples

			For n = 4 there are A000108(3) = 5 possible parenthesizations of x^x^x^x: [x^(x^(x^x)), x^((x^x)^x), (x^(x^x))^x, (x^x)^(x^x), ((x^x)^x)^x]. The first, second, third, fourth derivatives at x=1 are [1,1,1,1,1], [2,2,4,4,6], [9,15,18,18,27], [56,80,100,100,156] => row 4 = [1,3,4,4].
Triangle T(n,k) begins:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  4,  4;
  1, 4,  7,  9,  9;
  1, 5, 11, 17, 20,  20;
  1, 6, 15, 30, 45,  48,  48;
  1, 7, 20, 50, 92, 113, 115, 115;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..16, flattened
• Bradley Klee, Plot 1
• Bradley Klee, Plot 2

Crossrefs

Columns k=1-10 give: A000012, A028310, A199085, A199205, A199296, A199883, A215796, A215971, A216062, A216403.

Main diagonal gives (conjectured): A000081.

Cf. A000108, A215703.

Programs

Maple

with(combinat):
F:= proc(n) F(n):=`if`(n<2, [(x+1)$n], map(h->(x+1)^h, g(n-1, n-1))) end:
g:= proc(n, i) option remember; `if`(n=0 or i=1, [(x+1)^n],
     `if`(i<1, [], [seq(seq(seq(mul(F(i)[w[t]-t+1], t=1..j)*v,
      w=choose([$1..nops(F(i))+j-1], j)), v=g(n-i*j, i-1)), j=0..n/i)]))
    end:
T:= proc(n) local i, l;
      l:= map(f->[seq(i!*coeff(series(f, x, n+1), x, i), i=1..n)], F(n));
      seq(nops({map(x->x[i], l)[]}), i=1..n)
    end:
seq(T(n), n=1..10);

Mathematica

g[n_, i_] := g[n, i] = If[i==1, {x^n}, Flatten@Table[Table[Table[Product[ T[i][[w[[t]] - t+1]], {t, 1, j}]*v, {v, g[n - i*j, i-1]}], {w, Subsets[ Range[Length[T[i]] + j - 1], {j}]}], {j, 0, n/i}]];
T[n_] := T[n] = If[n==1, {x}, x^#& /@ g[n-1, n-1]];
T[n_, k_] := Union[k! (SeriesCoefficient[#, {x, 0, k}]& /@ (T[n] /. x -> x+1))] // Length;
Table[T[n, k], {n, 1, 11}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 08 2021, after Alois P. Heinz *)