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%I #11 Feb 11 2020 16:26:07
%S 1,-6,-180,-4200,-117600,-3969000,-115259760,-3030051024,
%T -106051785840,-3363975014400,-98227319533200,-3110531785218000,
%U -103492384705710000,-3211851661880141280,-96355549856404238400,-3303308595002047632000,-106681852579497947388000
%N Minimum value in the inverse of Hilbert's matrix.
%H T. D. Noe, <a href="/A210358/b210358.txt">Table of n, a(n) for n = 1..100</a>
%t Table[im = Inverse[HilbertMatrix[n]]; Min[Flatten[im]]], {n, 20}]
%o (PARI) a(n)=vecmin(1/mathilbert(n)) \\ _Hugo Pfoertner_, Feb 11 2020
%Y Cf. A061065, A189765 (inverse matrix), A210357.
%K sign
%O 1,2
%A _T. D. Noe_, Apr 02 2012
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