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.
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
Examples
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; ...
Programs
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PARI
{P(n,k,q=3) = my(A=Mat(1),B);if(nPaul D. Hanna, Jul 11 2025): for(n=0,10, for(k=0,n, print1(P(n,k),", ")); print(""))
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).
Comments