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Numerators of coefficients in power series for -log(1+x)*log(1-x).
6

%I #38 Jun 10 2018 17:55:56

%S 1,5,47,319,1879,20417,263111,52279,1768477,33464927,166770367,

%T 3825136961,19081066231,57128792093,236266661971,7313175618421,

%U 14606816124167,102126365345729,3774664307989373,3771059091081773

%N Numerators of coefficients in power series for -log(1+x)*log(1-x).

%H Muniru A Asiru, <a href="/A049281/b049281.txt">Table of n, a(n) for n = 1..300</a>

%F a(n)/A069685(n) = Integral_{x=0..1} x^(n-1)*log(1 + 1/sqrt(x)) dx = 1/n*Sum_{k=0..2*n-2} (-1)^k/(2*n-1-k).

%F From _Peter Bala_, Feb 21 2017: (Start)

%F a(n) = numerator((1/n)*Sum_{k=1..2*n-1} (-1)^(k-1)/k ). Cf. A058313.

%F a(n) = numerator((1/2)*binomial(2*n,n)*Sum_{k=0..n-1} (-1)^k* binomial(n-1,k)/(n + k)^2 ).

%F The coefficients in the expansion of log(1 + x)*log(1 - x) are given by (1/2)*binomial(2*n,n)*Integral_{x = 0..1} (x*(1 - x))^(n-1)*log(x) dx.

%F log(1 + x)*log(1 - x) = (1/2)*Integral_{z = 0..1} log(z)/(z*(1 - z)) * (1/sqrt( 1 - 4*x^2*z*(1 - z) ) - 1) dz. (End)

%e -log(1 + x)*log(1 - x) = x^2 + 5/12*x^4 + 47/180*x^6 + 319/1680*x^8 + ...

%p seq(numer(1/n*add( (-1)^k/(2*n-1-k),k = 0..2*n-2)), n = 1..20); # _Peter Bala_, Feb 21 2017

%o (GAP) List(List([1..25],n->(1/n)*Sum([1..2*n-1],k->(-1)^(k-1)/k)),NumeratorRat); # _Muniru A Asiru_, Jun 01 2018

%Y Bisection of A058313. Cf. A069685 (denominators).

%K easy,frac,nonn

%O 1,2

%A _Benoit Cloitre_, May 03 2002