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%I #31 Oct 10 2017 02:44:35
%S 4,44,428,10196,10719068,25865068,5472607916,74185965772,
%T 264698472181028,2290048394728148,19435959308462817284,
%U 2753151578548809148,20586893910854623222436,134344844535611780572028924
%N Numerator of the coefficient of x^(2n) in the expansion of 1/x^4 - 1/(3*x^2) - 1/(x^3*arctanh(x)).
%C 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].
%C 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.
%C 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
%C 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
%p A187870 := proc(n)
%p 1/x^4 -1/(3*x^2) -1/(x^3*arctanh(x)) ;
%p coeftayl(%,x=0,2*n) ;
%p numer(%) ;
%p end proc:
%p seq(A187870(n),n=0..10) ; # _R. J. Mathar_, Sep 21 2011
%p # Or
%p 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
%t 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 *)
%Y Cf. A195466 (denominator).
%K nonn,frac
%O 0,1
%A _Groux Roland_, Mar 14 2011
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