A309001 - OEIS

%I #26 Dec 09 2024 17:27:51

%S 3,3,6,4,60,840,2520,3465,360360,11440,2450448,23279256,46558512,

%T 356948592,2230928700,80313433200,2329089562800,144403552893600,

%U 3702655202400,72201776446800,1068586291412640,763275922437600,5215718803323600,40777437916893600,103511957789037600,718702343285249700

%N a(n) is the denominator of the rational part of Sum_{k>=n} binomial(2*k,k-n)^(-1).

%C The sum is a rational number plus an integer multiple of Pi/(9*sqrt(3)).

%H Robert Israel, <a href="/A309001/b309001.txt">Table of n, a(n) for n = 0..1154</a>

%H Mathematics StackExchange, <a href="https://math.stackexchange.com/questions/3253047/sum-of-reciprocal-binomial-coefficients/3253096#3253096">Sum of reciprocal binomial coefficients</a>

%F Sum_{k>=n} binomial(2*k,k-n)^(-1) = Integral_{t=0..1} (1-t)^(2*n)*(2+(2*n-1)*(1-t+t^2))/(1-t+t^2)^2 dt.

%F G.f. of the rational part is -(4 + x + 4*x^2)/(3*(-1 + x)*(1 + x + x^2)) - ((1 + 3*x + x^2)*log(1 - x)*x)/(2*(1 + x + x^2)^2) + 2*arctanh(sqrt(x))*(1 + x)*x^(3/2)/(1 + x + x^2)^2.

%e Sum_{k>=3} binomial(2*k,k-3)^(-1) = 3/4 + 2*Pi/(9*sqrt(3)) so a(3) = 4.

%p G:= -(4 + x + 4*x^2)/(3*(-1 + x)*(1 + x + x^2)) - ((1 + 3*x + x^2)*ln(1 - x)*x)/(2*(1 + x + x^2)^2) + 2*arctanh(sqrt(x))*(1 + x)*x^(3/2)/(1 + x + x^2)^2:

%p S:= series(G,x,101):

%p map(denom, [seq(coeff(S,x,i),i=0..100)]);

%t a[n_] := FunctionExpand[Sum[1/Binomial[2k, k-n], {k, n, Infinity}]] /. Pi -> 0 // Denominator;

%t a /@ Range[0, 40] (* _Jean-François Alcover_, Jul 31 2020 *)

%Y Cf. A308559 (numerators).

%K nonn,frac

%O 0,1

%A _Robert Israel_, Jul 04 2019