%I #5 Mar 30 2012 16:50:35
%S 1,1,2,3,12,20,300,525,1960,49392,1481760,5821200,164656800,336370320,
%T 3741185448
%N Sequence terms are generated by solving the n x n linear algebra problem [H]x = b, where b is the unit vector. Only xn, the last unknown is used.
%C These matrices are notorious for being illconditioned, they are solved using rational arithmetic. To reproduce these numbers, or larger ones, one can edit the program above with a text editor and then drag it into the Maxima window. The last term required approx. 10 minutes to generate on an Intel dualcore 6600 clocked at 2.4 GHz with 2 Gig of RAM. To change the order of the Hilbert matrix, just change the order variable at the top of the file. It is currently 4, which runs quickly. So a point of curiosity really. The numbers become extremely expensive to find, growing at least as the cube of n times a large constant. So to me they are a kind of symbolic gold. The 100th number for example, is probably not knowable at the current time, but that is speculation on my part. Perhaps you will observe some functional relationship that allows their simple generation thus "Cracking the HilbertWarren Sequence Code". For example 12 = 4 * 4  4, 20 = 5 * 5  5, 300 = 20 * 20  100, 525 = 25 * 25  100 and so on.
%H William Schelter, <a href="http://sourceforge.net/project/showfiles.php?group_id=4933">DOE Maxima  A Symbolic Algebra Program</a>
%H Gilbert Strang, <a href="http://www.amazon.com/LinearAlgebraApplicationsGilbertStrang/dp/0030105676/sr=11/qid=1161623501/ref=sr_1_1/00299855751193654?ie=UTF8&s=books#citebody">Linear Algebra and Its Applications</a>
%H L. V. Warren, <a href="http://www.blogger.com/posts.g?blogID=28407747">Cracking the HilbertWarren Code</a>
%F Solve the n x n linear algebra problem [H]x = b, where b is the unit vector and n is the order. Only xn, the last unknown is used.
%e Triangularization of [H] is performed, followed by inversion or back substitution. The terms are the denominators of the last solution xn to the Hilbert matrices of order 1, 2, ... 15, respectively. The numerator is unity in all cases.
%e For order 3 problem the solutions are: ([1/6],[ 1/3],[1/2]), thus a() = 2, since we just take the denominator.
%o load("eigen"); order : 4; X : columnvector(makelist(concat(x,i), i, 1, order)); h[i,j] := 1/(i + j 1); Unity[i,j] := 1; A : genmatrix(h, order, order); A . X; B : genmatrix(Unity, 1, order); A . X = B; Ap : triangularize(A); Ap . X = B; App : invert(Ap); App . B;
%Y Cf. A124262.
%K eigen,frac,hard,nonn
%O 1,3
%A L. Van Warren (van(AT)wdv.com), Oct 23 2006
%E The author said that this version of the submission was unsatisfactory and set me a better version in email, which may replace this.  _N. J. A. Sloane_, Oct 24 2006
