A116071 - OEIS

A116071

Triangle T, read by rows, equal to Pascal's triangle to the matrix power of Pascal's triangle, so that T = C^C, where C(n,k) = binomial(n,k) and T(n,k) = A000248(n-k)*C(n,k).

1, 1, 1, 3, 2, 1, 10, 9, 3, 1, 41, 40, 18, 4, 1, 196, 205, 100, 30, 5, 1, 1057, 1176, 615, 200, 45, 6, 1, 6322, 7399, 4116, 1435, 350, 63, 7, 1, 41393, 50576, 29596, 10976, 2870, 560, 84, 8, 1, 293608, 372537, 227592, 88788, 24696, 5166, 840, 108, 9, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

Column 0 = A000248 (Number of forests with n nodes and height at most 1).

Column 1 = A052512 (Number of labeled trees of height 2).

Row sums = A080108 (Sum_{k=1..n} k^(n-k) * C(n-1,k-1)).

Central terms = A116072(n) = (n+1) * A000108(n) * A000248(n).

From Peter Bala, Sep 13 2012: (Start)

For commuting lower unitriangular matrix A and lower triangular matrix B we define A raised to the matrix power B, denoted by A^B, to be the lower unitriangular matrix Exp(B*Log(A)). Here Exp denotes the matrix exponential defined by the power series

Exp(A) = 1 + A + A^2/2! + A^3/3! + ...

and the matrix logarithm Log(A) is defined by the series

Log(A) = (A-1) - 1/2*(A-1)^2/2 + 1/3*(A-1)^3 - ....

Let A = [f(x),x] and B = [g(x),x] be exponential Riordan arrays in the Appell subgroup and suppose f(0) = 1. Then A and B commute and A^B is the exponential Riordan array [exp(g(x)*log(f(x))),x], also belonging to the Appell group. In the present case we are taking A = B = [exp(x),x], equal to the Pascal triangle A007318.

For any lower unitriangular matrix A (with, say, rational entries) the infinite tower of powers A^(A^(A^...))) is well-defined (and also has rational entries). An example is given in the Formula section.

(End)

LINKS

Paul D. Hanna, Table of n, a(n) for n = 0..1080 (rows 0..45 of flattened triangle).

FORMULA

E.g.f.: exp( x*exp(x) + x*y ).

From Peter Bala, Sep 13 2012: (Start)

Exponential Riordan array [exp(x*exp(x)),x] belonging to the Appell group. Thus the e.g.f. for the k-th column of the triangle is x^k/k!*exp(x*exp(x)).

The inverse array, denote it by X, is a signed version of A215652. The infinite tower of matrix powers X^(X^(X^(...))) equals the inverse of Pascal's triangle.

(End)

O.g.f.: Sum_{n>=0} x^n / (1 - x*(n+y))^(n+1). - Paul D. Hanna, Aug 03 2014

G.f. for row n: Sum_{k=0..n} binomial(n,k) * (k + y)^(n-k) for n>=0. - Paul D. Hanna, Aug 03 2014

T(n,k) = Sum_{j=0..n-k} C(n,j) * C(n-j,k) * j^(n-k-j) = A000248(n-k)*C(n,k). - Paul D. Hanna, Aug 03 2014

Infinitesimal generator is A216973. - Peter Bala, Feb 13 2017

EXAMPLE

E.g.f.: E(x,y) = 1 + (1 + y)*x + (3 + 2*y + y^2)*x^2/2!

+ (10 + 9*y + 3*y^2 + y^3)*x^3/3!

+ (41 + 40*y + 18*y^2 + 4*y^3 + y^4)*x^4/4!

+ (196 + 205*y + 100*y^2 + 30*y^3 + 5*y^4 + y^5)*x^5/5! +...

where E(x,y) = exp(x*y) * exp(x*exp(x)).

O.g.f.: A(x,y) = 1 + (1 + y)*x + (3 + 2*y + y^2)*x^2

+ (10 + 9*y + 3*y^2 + y^3)*x^3

+ (41 + 40*y + 18*y^2 + 4*y^3 + y^4)*x^4

+ (196 + 205*y + 100*y^2 + 30*y^3 + 5*y^4 + y^5)*x^5 +...

where

A(x,y) = 1/(1-x*y) + x/(1-x*(y+1))^2 + x^2/(1-x*(y+2))^3 + x^3/(1-x*(y+3))^4 + x^4/(1-x*(y+4))^5 + x^5/(1-x*(y+5))^6 + x^6/(1-x*(y+6))^7 + x^7/(1-x*(y+7))^8 +...

Triangle begins:

1, 1;

3, 2, 1;

10, 9, 3, 1;

41, 40, 18, 4, 1;

196, 205, 100, 30, 5, 1;

1057, 1176, 615, 200, 45, 6, 1;

6322, 7399, 4116, 1435, 350, 63, 7, 1;

41393, 50576, 29596, 10976, 2870, 560, 84, 8, 1;

293608, 372537, 227592, 88788, 24696, 5166, 840, 108, 9, 1;

2237921, 2936080, 1862685, 758640, 221970, 49392, 8610, 1200, 135, 10, 1; ...

MATHEMATICA

(* The function RiordanArray is defined in A256893. *)

RiordanArray[Exp[# Exp[#]]&, #&, 10, True] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

PROG

(PARI) /* By definition C^C: */

{T(n, k)=local(A, C=matrix(n+1, n+1, r, c, binomial(r-1, c-1)), L=matrix(n+1, n+1, r, c, if(r==c+1, c))); A=sum(m=0, n, L^m*C^m/m!); A[n+1, k+1]}