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%I #23 Jul 31 2020 10:02:02
%S 4,1,23,3,-211,6169,1709,-24923,3941153,7457,-26565167,338662421,
%T 29719175,-5168552017,40526745521,50607208969,-42190362918239,
%U 3146154503067509,2312776975921,-1570173112141273,27153272350852367,473757364639811,-132365433369215539,1183965646415001041,63942535017037643
%N a(n) is the numerator 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="/A308559/b308559.txt">Table of n, a(n) for n = 0..1149</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) = int_0^1 dt (1-t)^(2*n)*(2+(2*n-1)*(1-t+t^2))/(1-t+t^2)^2.
%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) = 3.
%p f:= proc(n) local J;
%p J:= int((1-t)^(2*n)*(2+(2*n-1)*(1-t+t^2))/(1-t+t^2)^2,t=0..1);
%p numer(subs(Pi=0, J))
%p end proc:
%p map(f, [$0..40]);
%t a[n_] := FunctionExpand[Sum[1/Binomial[2k, k-n], {k, n, Infinity}]] /. Pi -> 0 // Numerator;
%t a /@ Range[0, 40] (* _Jean-François Alcover_, Jul 31 2020 *)
%Y Cf. A309001 (denominators).
%K sign,frac
%O 0,1
%A _Robert Israel_, Jun 07 2019
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