OFFSET
0,1
COMMENTS
Let f(x) = 1/x^4 - 1/(3*x^2) - 1/(x^3*arctanh(x)) = Sum_{n>=0} r(n)*x^(2n), then a(n) is the numerator of r(n), and r(n) is also the moment of order n for the density rho(x) = 2*sqrt(x)/(4*(arctanh(sqrt(x)))^2 + Pi^2) over the interval [0,1].
r(n) can also be evaluated as (-1)^(n+1)*det(An) with An the square matrix of order n+2 defined by: if j <= i A[i,j] = 1/(2*i-2*j+3), A[i,i+1]=1, if j > i+1 A[i,j]=0.
A very similar sequence of numerators 1, 1, 4, 44, 428, 10196, ... (from there on apparently the same as here) is constructed from the fractions c(0)=-1 and c(n) = Sum_{i=0..n-1} c(i)/(2n-2i+1), which is c(0)=-1, c(1)=1/3, c(2)=4/45, c(3)= 44/945, etc. The recurrence is designed to ensure that Sum_{i=0..n} c(i)/(2n-2i+1) = 0. - Paul Curtz, Sep 15 2011
Prepending 1 to the data gives the (-1)^n times the numerator of the odd powers in the expansion of 1/arctan(x). - Peter Luschny, Oct 04 2014
MAPLE
A187870 := proc(n)
1/x^4 -1/(3*x^2) -1/(x^3*arctanh(x)) ;
coeftayl(%, x=0, 2*n) ;
numer(%) ;
end proc:
seq(A187870(n), n=0..10) ; # R. J. Mathar, Sep 21 2011
# Or
seq((-1)^n*numer(coeff(series(1/arctan(x), x, 2*n+2), x, 2*n+1)), n=1..14); # Peter Luschny, Oct 04 2014
MATHEMATICA
a[n_] := Sum[(2^(j+1)*Binomial[2*n+3, j]*Sum[(k!*StirlingS1[j+k, j]*StirlingS2[j+1, k])/(j+k)!, {k, 0, j+1}])/(j+1), {j, 0, 2*n+3}]/ (2*n+3); Table[a[n] // Numerator, {n, 0, 13}] (* Jean-François Alcover, Jul 03 2013, after Vladimir Kruchinin's formula in A216272 *)
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Groux Roland, Mar 14 2011
STATUS
approved