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%I #21 Oct 11 2021 18:58:56
%S 1,3,10,105,252,2310,25740,9009,136136,11639628,10581480,223092870,
%T 1029659400,2868336900,11090902680,644658718275,606737617200,
%U 4011209802600,140603459396400,133573286426580,5215718803323600
%N Denominator of the sum of all elements in the n X n Hilbert matrix M(i,j) = 1/(i+j-1), where i,j = 1..n.
%C Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) = A117731(n) / A117664(n) = 2n * H'(2n) = 2n * A058313(2n) / A058312(2n), where H'(2n) is 2n-th alternating sign Harmonic Number. H'(2n) = H(2n) - H(n), where H(n) is n-th Harmonic Number. - _Alexander Adamchuk_, Apr 23 2006
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HilbertMatrix.html">Hilbert Matrix</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicNumber.html">Harmonic Number</a>
%F a(n) = A111876(n-1)/n.
%F a(n) = denominator( Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) ). Numerator is A117731(n). - _Alexander Adamchuk_, Apr 23 2006
%F a(n) = denominator( Sum_{k=1..n} (2*k)/(n+k) ). - _Peter Bala_, Oct 10 2021
%e For n=2, the 2 X 2 Hilbert matrix is [1, 1/2; 1/2, 1/3], so a(2) = denominator(1 + 1/2 + 1/2 + 1/3) = denominator(7/3) = 3.
%e The n X n Hilbert matrix begins:
%e 1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
%e 1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
%e 1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
%e 1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
%e 1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
%e 1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
%e ...
%t Table[Denominator[Sum[1/(i + j - 1), {i, n}, {j, n}]], {n, 30}]
%Y Cf. A091342, A098118, A111876, A082687, A086881, A005249, A001008, A002805.
%Y Numerator is A117731(n).
%K nonn
%O 1,2
%A _Alexander Adamchuk_, Apr 11 2006
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