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%I #11 Feb 28 2022 12:11:07
%S 1,10,70,420,4620,60060,60060,408408,7759752,38798760,892371480,
%T 4461857400,13385572200,55454513400,1719089915400,3438179830800,
%U 24067258815600,890488576177200,890488576177200,36510031623265200,1569931359800403600,1569931359800403600,73786773910618969200
%N Denominator of partial sums of a certain series.
%C The numerators are given in A101028.
%C One third of the denominator of the finite differences of the series of sums of all matrix elements of n X n Hilbert matrix M(i,j)=1/(i+j-1) (i,j = 1..n). - _Alexander Adamchuk_, Apr 11 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) = denominator(s(n)) with s(n)=3*sum(1/((2*k-1)*k*(2*k+1)), k=1..n). See A101028 for more information.
%F a(n) = 1/3*Denominator[Sum[Sum[1/(i+j-1),{i,1,n+1}],{j,1,n+1}]-Sum[Sum[1/(i+j-1),{i,1,n}],{j,1,n}]]. a(n) = 1/3*Denominator[H(2n+1) + H(2n) - 2H(n)], where H(n) = Sum[1/k, (k, 1, n}] is a Harmonic number, H[n] = A001008/A002805. - _Alexander Adamchuk_, Apr 11 2006
%e n=2: HilbertMatrix[n,n]
%e 1 1/2
%e 1/2 1/3
%e so a(1) = 1/3*Denominator[(1 + 1/2 + 1/2 + 1/3) - 1] = 1/3*Denominator[7/3 -1] = 1/3*Denominator[4/3] = 1.
%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 ...
%t Denominator[Table[Sum[1/(i + j - 1), {i, n}, {j, n}], {n,2, 27}]-Table[Sum[1/(i + j - 1), {i, n}, {j, n}], {n, 26}]]/3 (* _Alexander Adamchuk_, Apr 11 2006 *)
%o (PARI) a(n) = denominator(3*sum(k=1, n, 1/((2*k-1)*k*(2*k+1)))); \\ _Michel Marcus_, Feb 28 2022
%Y Cf. A098118, A086881, A005249, A001008, A002805.
%Y Cf. A101028 (numerators).
%K nonn,frac,easy
%O 1,2
%A _Wolfdieter Lang_, Dec 17 2004
%E More terms from _Michel Marcus_, Feb 28 2022
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