|
|
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!.
|
|
8
|
|
|
0, 1, -1, 1, -2, 4, -12, 36, -144, 576, -2880, 14400, -86400, 518400, -3628800, 25401600, -203212800, 1625702400, -14631321600, 131681894400, -1316818944000, 13168189440000, -144850083840000, 1593350922240000, -19120211066880000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,5
|
|
COMMENTS
|
Signed version of A010551, with leading zero.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (-1)^(n-1) * floor((n-1)/2)! * floor(n/2)! for n > 0, with a(0)=0.
E.g.f.: A(x) = (1-x/2)/sqrt(1-x^2/4)*arccos(1-x^2/2).
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
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
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
Conjecture: 4*a(n) + 2*a(n-1) - (n-1)*(n-2)*a(n-2) = 0, n > 2. - R. J. Mathar, Nov 25 2015
|
|
EXAMPLE
|
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.
|
|
PROG
|
(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]))}
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|