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%I #6 Jun 13 2017 22:36:32
%S 1,1,1,3,3,1,18,18,9,1,216,216,135,27,1,5589,5589,4050,1134,81,1,
%T 336555,336555,269730,95256,9963,243,1,49768101,49768101,42724503,
%U 17926839,2450898,88938,729,1,18707873562,18707873562,16835895603,8074043145
%N Triangle P, read by rows, that satisfies [P^3](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(3*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(k,k)=1 and P(k+1,1)=P(k+1,0) for k>=0.
%C Column 0 and column 1 are equal for n>0.
%F Let q=3; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x = -Sum_{n>=1} Product_{j=0..n-1} -L(q^j*x)/(j+1); L(x) equals the g.f. of column 0 of the matrix log of P (A111844).
%e Let q=3; the g.f. of column k of matrix power P^m is:
%e 1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
%e (m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
%e (m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
%e where L(x) satisfies:
%e x = L(x) - L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! -+ ...
%e and L(x) = x + 3/2!*x^2 + 27/3!*x^3 + 486/4!*x^4 +...(A111844).
%e Thus the g.f. of column 0 of matrix power P^m is:
%e 1 + m*L(x) + m^2/2!*L(x)*L(3*x) + m^3/3!*L(x)*L(3*x)*L(3^2*x) +
%e m^4/4!*L(x)*L(3*x)*L(3^2*x)*L(3^3*x) + ...
%e Triangle P begins:
%e 1;
%e 1,1;
%e 3,3,1;
%e 18,18,9,1;
%e 216,216,135,27,1;
%e 5589,5589,4050,1134,81,1;
%e 336555,336555,269730,95256,9963,243,1; ...
%e where P^3 shifts columns left and up one place:
%e 1;
%e 3,1;
%e 18,9,1;
%e 216,135,27,1;
%e 5589,4050,1134,81,1; ...
%o (PARI) P(n,k,q=3)=local(A=Mat(1),B);if(n<k || k<0,0, for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,if(j==1,B[i,j]=(A^q)[i-1,1], B[i,j]=(A^q)[i-1,j-1]));));A=B);return(A[n+1,k+1]))
%Y Cf. A111841 (column 0), A111842 (row sums), A111843 (matrix log), A078122 (variant).
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Aug 22 2005
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