OFFSET
0,5
COMMENTS
Also P(n,k) = the partitions of (5^n - 5^(n-k)) into powers of 5 <= 5^(n-k).
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).
EXAMPLE
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; ...
PROG
(PARI) P(n, k, q=5)=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 || j==1, B[i, j]=1, B[i, j]=(A^q)[i-1, j-1]); )); A=B); return(A[n+1, k+1]))
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gottfried Helms and Paul D. Hanna, Aug 22 2005
STATUS
approved
