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a(n) = (2*n)!*Sum[Sum[1/(i+j),{i,1,n}],{j,1,n}]
6

%I #9 May 10 2013 12:45:25

%S 1,34,1788,146256,17485920,2894002560,635331029760,178910029670400,

%T 62920533840998400,27042268338763776000,13950701922125574144000,

%U 8509745665997194493952000,6059691013778107566981120000

%N a(n) = (2*n)!*Sum[Sum[1/(i+j),{i,1,n}],{j,1,n}]

%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) = (2*n)!*((2*n+1)*Psi(2*n+2)-(2*n+2)*Psi(n+2)+1-gamma). limit(a(n)/(2*n)!/n, n=infinity)=2*ln2. - _Vladeta Jovovic_, Aug 24 2003

%F Sum of all matrix elements M(i, j) = 1/(i+j) multiplied by (2*n)! (i, j = 1..n) or Sum of all matrix elements M(i, j) = 2*i/(i+j)^2 multiplied by (2*n)! (i, j = 1..n). a(n) = (2*n)!*Sum[Sum[2*i/(i+j)^2, {i, 1, n}], {j, 1, n}] - _Alexander Adamchuk_, Oct 24 2004

%F a(n) = (2n)! * ((2n+2)*H(2n+2) - 2(n+1)*H(n+1) - H(2n+1)), where H(n) is HarmonicNumber[n] = Sum[1/i, {i, 1, n}] = A001008(n)/A002805(n). - _Alexander Adamchuk_, Nov 01 2004

%e a(2) = 4!*(1/(1+1)+1/(1+2)+1/(2+1)+1/(2+2)) = 24*(1/2+1/3+1/3+1/4)

%t Table[((2*n)!)*Sum[Sum[1/(a+b), {i, 1, n}], {j, 1, n}], {n, 1, 20}]

%Y Cf. A098118, A001008, A002805.

%K nonn

%O 1,2

%A _Alexander Adamchuk_, Aug 21 2003