cp's OEIS Frontend

A111824 Column 0 of the matrix logarithm (A111823) of triangle A111820, which shifts columns left and up under matrix 5th power; these terms are the result of multiplying the element in row n by n!.

%I A111824 #7 Mar 13 2015 19:21:58
%S A111824 0,1,-3,16,2814,-1092180,-3603928080,58978973128440,
%T A111824 5974833380453777520,-3294186866481455009752320,
%U A111824 -10279982482873484428390722523200,175129088125361734252730927280177244800
%N A111824 Column 0 of the matrix logarithm (A111823) of triangle A111820, which shifts columns left and up under matrix 5th power; these terms are the result of multiplying the element in row n by n!.
%C A111824 Let q=5; the g.f. of column k of A111820^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).
%F A111824 E.g.f. satisfies: x/(1-x) = Sum_{n>=1} Prod_{j=0..n-1} A(5^j*x)/(j+1).
%e A111824 A(x) = x - 3/2!*x^2 + 16/3!*x^3 + 2814/4!*x^4 - 1092180/5!*x^5 +...
%e A111824 where e.g.f. A(x) satisfies:
%e A111824 x/(1-x) = A(x) + A(x)*A(5*x)/2! + A(x)*A(5*x)*A(5^2*x)/3! +
%e A111824 A(x)*A(5*x)*A(5^2*x)*A(5^3*x)/4! + ...
%e A111824 Let G(x) be the g.f. of A111821 (column 1 of A111820), then
%e A111824 G(x) = 1 + 5*A(x) + 5^2*A(x)*A(5*x)/2! +
%e A111824 5^3*A(x)*A(5*x)*A(5^2*x)/3! +
%e A111824 5^4*A(x)*A(5*x)*A(5^2*x)*A(5^3*x)/4! + ...
%o A111824 (PARI) {a(n,q=5)=local(A=x/(1-x+x*O(x^n)));for(i=1,n, A=x/(1-x)/(1+sum(j=1,n,prod(k=1,j,subst(A,x,q^k*x))/(j+1)!))); return(n!*polcoeff(A,n))}
%Y A111824 Cf. A111820 (triangle), A111821, A111823 (matrix log); A110505 (q=-1), A111814 (q=2), A111816 (q=3), A111819 (q=4), A111829 (q=6), A111834 (q=7), A111839 (q=8).
%K A111824 sign
%O A111824 0,3
%A A111824 _Gottfried Helms_ and _Paul D. Hanna_, Aug 22 2005