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%I #26 Jun 19 2022 13:42:36
%S 1,4,12,9,192,180,16,1200,6480,2800,25,4800,79380,179200,44100,36,
%T 14700,564480,3628800,4410000,698544,49,37632,2857680,40320000,
%U 133402500,100590336,11099088,64,84672,11430720,304920000,2134440000,4249941696,2175421248,176679360
%N Triangular table read by rows: T(n,k) is the k-th entry of the main diagonal of the inverse Hilbert matrix of order n.
%H Jianing Song, <a href="/A348419/b348419.txt">Table of n, a(n) for n = 1..5050</a> (first 100 rows)
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HilbertMatrix.html">Hilbert Matrix</a>
%e The inverse Hilbert matrix of order 4 is given by
%e [ 16 -120 240 -140]
%e [-120 1200 -2700 1680]
%e [ 240 -2700 6480 -4200]
%e [-140 1680 -4200 2800].
%e Hence the 4th row is 16, 1200, 6480, 2800.
%e The first 8 rows of the table are:
%e 1,
%e 4, 12,
%e 9, 192, 180,
%e 16, 1200, 6480, 2800,
%e 25, 4800, 79380, 179200, 44100,
%e 36, 14700, 564480, 3628800, 4410000, 698544,
%e 49, 37632, 2857680, 40320000, 133402500, 100590336, 11099088,
%e 64, 84672, 11430720, 304920000, 2134440000, 4249941696, 2175421248, 176679360,
%e ...
%p T:= n-> (M-> seq(M[i, i], i=1..n))(1/LinearAlgebra[HilbertMatrix](n)):
%p seq(T(n), n=1..8); # _Alois P. Heinz_, Jun 19 2022
%t T[n_, k_] := Inverse[HilbertMatrix[n]][[k, k]]; Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* _Amiram Eldar_, Oct 18 2021 *)
%o (PARI) T(n,k) = (1/mathilbert(n))[k,k]
%Y Cf. A189766 (row sums), A189765, A005249.
%Y A210356 gives the maximum value of each row and A210357 gives the positions of the maximum values.
%Y Main diagonal gives A000515(n-1).
%K nonn,tabl
%O 1,2
%A _Jianing Song_, Oct 18 2021