A111845 Triangle P, read by rows, that satisfies [P^4](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(4*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, 4, 4, 1, 40, 40, 16, 1, 1040, 1040, 544, 64, 1, 78240, 78240, 48960, 8320, 256, 1, 18504256, 18504256, 13110400, 2878720, 131584, 1024, 1, 14463224448, 14463224448, 11192599808, 2982187520, 180270080, 2099200, 4096, 1
Offset: 0
Examples
Let q=4; 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 + 4/2!*x^2 + 56/3!*x^3 + 1728/4!*x^4 + ... (A111849). Thus the g.f. of column 0 of matrix power P^m is: 1 + m*L(x) + m^2/2!*L(x)*L(4*x) + m^3/3!*L(x)*L(4*x)*L(4^2*x) + m^4/4!*L(x)*L(4*x)*L(4^2*x)*L(4^3*x) + ... Triangle P begins: 1; 1,1; 4,4,1; 40,40,16,1; 1040,1040,544,64,1; 78240,78240,48960,8320,256,1; 18504256,18504256,13110400,2878720,131584,1024,1; ... where P^4 shifts columns left and up one place: 1; 4,1; 40,16,1; 1040,544,64,1; 78240,48960,8320,256,1; ...
Crossrefs
Programs
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PARI
{P(n,k,q=4) = 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=4; 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} -(-1)^n/n!*Product_{j=0..n-1} L(q^j*x); L(x) equals the g.f. of column 0 of the matrix log of P (A111849).
Comments