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A111840
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.
6
1, 1, 1, 3, 3, 1, 18, 18, 9, 1, 216, 216, 135, 27, 1, 5589, 5589, 4050, 1134, 81, 1, 336555, 336555, 269730, 95256, 9963, 243, 1, 49768101, 49768101, 42724503, 17926839, 2450898, 88938, 729, 1, 18707873562, 18707873562, 16835895603, 8074043145
OFFSET
0,4
COMMENTS
Column 0 and column 1 are equal for n>0.
FORMULA
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).
EXAMPLE
Let q=3; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x = L(x) - L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! -+ ...
and L(x) = x + 3/2!*x^2 + 27/3!*x^3 + 486/4!*x^4 +...(A111844).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(3*x) + m^3/3!*L(x)*L(3*x)*L(3^2*x) +
m^4/4!*L(x)*L(3*x)*L(3^2*x)*L(3^3*x) + ...
Triangle P begins:
1;
1,1;
3,3,1;
18,18,9,1;
216,216,135,27,1;
5589,5589,4050,1134,81,1;
336555,336555,269730,95256,9963,243,1; ...
where P^3 shifts columns left and up one place:
1;
3,1;
18,9,1;
216,135,27,1;
5589,4050,1134,81,1; ...
PROG
(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]))
CROSSREFS
Cf. A111841 (column 0), A111842 (row sums), A111843 (matrix log), A078122 (variant).
Sequence in context: A104417 A121438 A108391 * A174031 A228859 A259876
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Aug 22 2005
STATUS
approved