%I #11 Feb 11 2020 12:03:23
%S 1,12,192,6480,179200,4410000,133402500,4249941696,122367445200,
%T 3480673996800,117643011932160,3659449159080000,106518477825760000,
%U 3521767173114190000,114708987924290760000,3525270042097046880000,110552468520163390156800
%N Maximum modulus in the inverse of Hilbert's matrix.
%C This sequence corrects A061065, which was wrong for n > 6. These values appear at the A210357(n) position on the diagonal. Thanks to Clark Kimberling for verifying this sequence.
%H T. D. Noe, <a href="/A210356/b210356.txt">Table of n, a(n) for n = 1..100</a>
%t Table[im = Inverse[HilbertMatrix[n]]; Max[Abs[Flatten[im]]]], {n, 20}]
%o (PARI) for(n=1,17, my(h=1/mathilbert(n),s=0); for(j=1,n, for(k=1,n, s=max(s,h[j,k]))); print1(s,", ")) \\ _Hugo Pfoertner_, Feb 11 2020
%Y Cf. A061065, A189765 (inverse matrix), A210357, A210358 (minimum).
%K nonn
%O 1,2
%A _T. D. Noe_, Mar 28 2012