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Column 0 of the matrix logarithm (A111813) of triangle A078121, which shifts columns left and up under matrix square; these terms are the result of multiplying the element in row n by n!.
10

%I #7 Mar 30 2012 18:36:50

%S 0,1,0,-2,0,216,0,-568464,0,36058658688,0,-53694310935340800,0,

%T 1790669979087018171448320,0,-1280832788659041410080025283840000,0,

%U 18961468161294510864200732026858464699187200,0

%N Column 0 of the matrix logarithm (A111813) of triangle A078121, which shifts columns left and up under matrix square; these terms are the result of multiplying the element in row n by n!.

%C Surprisingly, the e.g.f. A(x) is an odd function: A(-x) = -A(x). Let q=2; the g.f. of column k of A078121^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).

%H Paul D. Hanna, <a href="/A111814/b111814.txt">Table of n, a(n), n== 0..50.</a>

%F E.g.f. satisfies: x/(1-x) = Sum_{n>=1} Prod_{j=0..n-1} A(2^j*x)/(j+1). E.g.f. satisfies: x/(1-x^2) = Sum_{n>=1}Prod_{j=0..2*n}A(2^j*x)/(j+1).

%e E.g.f.: A(x) = x - 2/3!*x^3 + 216/5!*x^5 - 568464/7!*x^7 + ...

%e where A(x) satisfies:

%e x/(1-x) = A(x) + A(x)*A(2*x)/2! + A(x)*A(2*x)*A(2^2*x)/3!

%e + A(x)*A(2*x)*A(2^2*x)*A(2^3*x)/4! + ...

%e also:

%e x/(1-x^2) = A(x) + A(x)*A(2*x)*A(2^2*x)/3!

%e + A(x)*A(2*x)*A(2^2*x)*A(2^3*x)*A(2^4*x)/5! + ...

%e Let G(x) be the g.f. of A002577 (column 1 of A078121), then

%e G(x) = 1 + 2*A(x) + 2^2*A(x)*A(2*x)/2! +

%e 2^3*A(x)*A(2*x)*A(2^2*x)/3! +

%e 2^4*A(x)*A(2*x)*A(2^2*x)*A(2^3*x)/4! + ...

%o (PARI) {a(n,q=2)=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 Cf. A078121 (triangle), A002577, A111813 (matrix log); A110505 (q=-1), A111816 (q=3), A111819 (q=4), A111824 (q=5), A111829 (q=6), A111834 (q=7), A111839 (q=8).

%K sign

%O 0,4

%A _Gottfried Helms_ and _Paul D. Hanna_, Aug 22 2005