|
| |
|
|
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.
|
|
10
|
|
|
|
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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,5
|
|
|
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).
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)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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;
...
|
|
|
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;
|
|
|
CROSSREFS
|
Main diagonal gives (conjectured): A000081.
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
