A111820 Triangle P, read by rows, that satisfies [P^5](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(5*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(0,k)=1 and P(k,k)=1 for all k>=0.
1, 1, 1, 1, 5, 1, 1, 55, 25, 1, 1, 2055, 1525, 125, 1, 1, 291430, 311525, 38875, 625, 1, 1, 165397680, 239305275, 40338875, 975625, 3125, 1, 1, 390075741430, 735920617775, 157056792000, 5077475625, 24409375, 15625, 1
Offset: 0
Examples
Let q=5; 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/(1-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 + 16/3!*x^3 + 2814/4!*x^4 +... (A111824). Thus the g.f. of column 0 of matrix power P^m is: 1 + m*L(x) + m^2/2!*L(x)*L(5*x) + m^3/3!*L(x)*L(5*x)*L(5^2*x) + m^4/4!*L(x)*L(5*x)*L(5^2*x)*L(5^3*x) + ... Triangle P begins: 1; 1,1; 1,5,1; 1,55,25,1; 1,2055,1525,125,1; 1,291430,311525,38875,625,1; 1,165397680,239305275,40338875,975625,3125,1; ... where P^5 shifts columns left and up one place: 1; 5,1; 55,25,1; 2055,1525,125,1; 291430,311525,38875,625,1; ...
Crossrefs
Programs
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PARI
P(n,k,q=5)=local(A=Mat(1),B);if(n
Formula
Let q=5; 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/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111824).
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