%I A107254
%S A107254 1,1,12,8640,870912000,22122558259200000,222531556847250309120000000,
%T A107254 1280394777025250130271722799104000000000,
%U A107254 5746332926632566442385615219551212618645504000000000000
%N A107254 SF(2n-1)/SF(n-1)^2 where SF is the superfactorial A000178 product of 
               factorials.
%C A107254 Inverse product of all matrix elements of n X n Hilbert Matrix M(i,j) 
               = 1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk (alex(AT)kolmogorov.com), 
               Apr 12 2006
%H A107254 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               HilbertMatrix.html">Link to a section of The World of Mathematics</
               a>.
%F A107254 a(n) = n!*(n+1)!*(n+2)!*...*(2n-1)!/(0!*1!*2!*3!*...*(n-1)!) = A000178(2n-1)/
               A000178(n-1)^2 = A079478(n)/A000984(n) = A079478(n-1)*A009445(n-1) 
               = A107252(n)*A000142(n) = A088020(n)/A039622(n).
%F A107254 a(n) = 1/Product[Product[1/(i+j-1),{i,1,n}],{j,1,n}]. - Alexander Adamchuk 
               (alex(AT)kolmogorov.com), Apr 12 2006
%e A107254 a(3) = 1!*2!*3!*4!*5!/(1!*2!*1!*2!) = 34560/4 = 8640.
%e A107254 n = 2: HilbertMatrix[n,n]
%e A107254 1 1/2
%e A107254 1/2 1/3
%e A107254 so a(2) = 1 / (1 * 1/2 * 1/2 * 1/3) = 12.
%e A107254 The n X n Hilbert matrix begins:
%e A107254 1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
%e A107254 1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
%e A107254 1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
%e A107254 1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
%e A107254 1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
%e A107254 1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
%t A107254 Table[Product[Product[(i+j-1),{i,1,n}],{j,1,n}],{n,1,10}] - Alexander 
               Adamchuk (alex(AT)kolmogorov.com), Apr 12 2006
%Y A107254 Cf. A002457.
%Y A107254 Cf. A098118, A005249.
%Y A107254 Sequence in context: A013481 A013513 A013480 this_sequence A012532 A012732 
               A001322
%Y A107254 Adjacent sequences: A107251 A107252 A107253 this_sequence A107255 A107256 
               A107257
%K A107254 nonn
%O A107254 0,3
%A A107254 Henry Bottomley (se16(AT)btinternet.com), May 14 2005