cp's OEIS Frontend

A111942 Column 0 of the matrix logarithm (A111941) of triangle A111940, which shifts columns left and up under matrix inverse; these terms are the result of multiplying the element in row n by n!.

%I A111942 #25 Jan 17 2018 02:56:49
%S A111942 0,1,-1,1,-2,4,-12,36,-144,576,-2880,14400,-86400,518400,-3628800,
%T A111942 25401600,-203212800,1625702400,-14631321600,131681894400,
%U A111942 -1316818944000,13168189440000,-144850083840000,1593350922240000,-19120211066880000
%N A111942 Column 0 of the matrix logarithm (A111941) of triangle A111940, which shifts columns left and up under matrix inverse; these terms are the result of multiplying the element in row n by n!.
%C A111942 Signed version of A010551, with leading zero.
%F A111942 a(n) = (-1)^(n-1) * floor((n-1)/2)! * floor(n/2)! for n > 0, with a(0)=0.
%F A111942 E.g.f.: A(x) = (1-x/2)/sqrt(1-x^2/4)*arccos(1-x^2/2).
%F A111942 G.f.: x*G(0) where G(k) = 1 - (k+1)*x/(1 - x*(k+1)/(x*(k+1) - 1/G(k+1) )); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Nov 28 2012
%F A111942 G.f.: G(0)*x/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(1*k+1) - 1/(1 + 1/(1 - x*(k+1)/(x*(1*k+1) - 1/G(k+1)))))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 20 2013
%F A111942 G.f.: x/G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/(1 - x*(k+1)/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 07 2013
%F A111942 Conjecture: 4*a(n) + 2*a(n-1) - (n-1)*(n-2)*a(n-2) = 0, n > 2. - _R. J. Mathar_, Nov 25 2015
%e A111942 E.g.f.: A(x) = x - (1/2!)*x^2 + (1/3!)*x^3 - (2/4!)*x^4 + (4/5!)*x^5 - (12/6!)*x^6 + (36/7!)*x^7 - (144/8!)*x^8 + (576/9!)*x^9 + ... where A(x)*A(-x) = -arccos(1-x^2/2)^2.
%o A111942 (PARI) {a(n,q=-1)=local(A=Mat(1),B);if(n<0,0, for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,if(j==1,B[i,j]=(A^q)[i-1,1], B[i,j]=(A^q)[i-1,j-1]));));A=B); B=sum(i=1,#A,-(A^0-A)^i/i);return(n!*B[n+1,1]))}
%Y A111942 Cf. A111940 (triangle), A111941 (matrix log), A110505 (variant), A010551 (unsigned).
%K A111942 sign
%O A111942 0,5
%A A111942 _Paul D. Hanna_, Aug 23 2005