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A098118
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Sum of all matrix elements of n X n Hilbert matrix M(i,j)=1/(i+j-1) (i,j = 1..n) multiplled by (2*n-1)!/n!: a(n) = (2*n-1)!/n!*Sum[Sum[1/(i+j-1),{i,1,n}],{j,1,n}].
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7
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1, 7, 74, 1066, 19524, 434568, 11393808, 343976400, 11752855200, 448372820160, 18892607771520, 871406506494720, 43669963405555200, 2362804077652300800, 137275789612950374400, 8523776656311156172800, 563309040416875548364800
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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LINKS
| Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
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FORMULA
| a(n) = (2*n-1)!/n!*Sum[Sum[1/(i+j-1), {i, 1, n}], {j, 1, n}]
a(n) = 2*(2*n-1)!/(n-1)!*H'(2n), where H'(2n) = H(2n) - H(n), H'(n) = Sum[(1/k)*(-1)^(k+1), (k, 1, n}] is an alternate signs Harmonic number, H(n) = Sum[1/k, (k, 1, n}] is a Harmonic number, H[n] = A001008/A002805. - Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 25 2004
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EXAMPLE
| n=2: HilbertMatrix[n,n]
1 1/2
1/2 1/3
so a(2) = (2*2-1)! / 2! * (1 + 1/2 + 1/2 + 1/3) = 7.
The n X n Hilbert matrix begins:
1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
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MATHEMATICA
| Table[(2n - 1)!/n! Sum[1/(i + j - 1), {i, n}, {j, n}], {n, 17}]
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CROSSREFS
| Cf. A086881, A005249, A001008, A002805.
Sequence in context: A137141 A114472 A000901 * A097821 A054745 A197091
Adjacent sequences: A098115 A098116 A098117 * A098119 A098120 A098121
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KEYWORD
| nonn
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AUTHOR
| Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 25 2004
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