login
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).
4

%I #31 Jul 19 2019 14:26:32

%S 1,1,1,3,2,1,10,9,3,1,41,40,18,4,1,196,205,100,30,5,1,1057,1176,615,

%T 200,45,6,1,6322,7399,4116,1435,350,63,7,1,41393,50576,29596,10976,

%U 2870,560,84,8,1,293608,372537,227592,88788,24696,5166,840,108,9,1

%N 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).

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

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

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

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

%C From _Peter Bala_, Sep 13 2012: (Start)

%C 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

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

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

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

%C 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.

%C 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.

%C (End)

%H Paul D. Hanna, <a href="/A116071/b116071.txt">Table of n, a(n) for n = 0..1080 (rows 0..45 of flattened triangle).</a>

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

%F From _Peter Bala_, Sep 13 2012: (Start)

%F 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)).

%F 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.

%F (End)

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

%F 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

%F 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

%F Infinitesimal generator is A216973. - _Peter Bala_, Feb 13 2017

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

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

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

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

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

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

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

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

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

%e where

%e 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 +...

%e Triangle begins:

%e 1;

%e 1, 1;

%e 3, 2, 1;

%e 10, 9, 3, 1;

%e 41, 40, 18, 4, 1;

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

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

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

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

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

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

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

%t RiordanArray[Exp[# Exp[#]]&, #&, 10, True] // Flatten (* _Jean-François Alcover_, Jul 19 2019 *)

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

%o {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]}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%o (PARI) /* From e.g.f.: */

%o {T(n,k)=local(A=1);A=exp( x*y + x*exp(x +x*O(x^n)) );n!*polcoeff(polcoeff(A, n,x),k,y)}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%o (PARI) /* From o.g.f. (_Paul D. Hanna_, Aug 03 2014): */

%o {T(n,k)=local(A=1);A=sum(k=0, n, x^k/(1 - x*(k+y) +x*O(x^n))^(k+1));polcoeff(polcoeff(A, n,x),k,y)}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%o (PARI) /* From row polynomials (_Paul D. Hanna_, Aug 03 2014): */

%o {T(n,k)=local(R);R=sum(k=0,n,(k+y)^(n-k)*binomial(n,k));polcoeff(R,k,y)}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%o (PARI) /* From formula for T(n,k) (_Paul D. Hanna_, Aug 03 2014): */

%o {T(n,k) = sum(j=0,n-k, binomial(n,j) * binomial(n-j,k) * j^(n-k-j))}

%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

%Y Cf. A000248, A052512, A080108, A116072, A215652.

%Y Cf. A080108, A216689, A240165, A245834, A245835, A216973.

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Feb 03 2006