number of partitions bench mark
Richard Mathar
mathar at strw.leidenuniv.nl
Wed May 14 10:19:49 CEST 2008
On Tue, May 13, 2008 at 12:12 PM, Jack Brennen <jb at brennen.net> wrote:
> Take the first 2*N integers and using each integer once and only
> once as either a numerator or a denominator, construct N fractions
> whose sum is an integer.
>
> How many unique solutions are there for each N?
>
> The sequence seems to begin: 1,1,7,21,190,...
Here are a couple of further values:
1, 1, 7, 21, 190, 1007, 6972,
I've got them with the following straightforward PARI/GP
implementation (taking about (2*n)!/n! iterations):
{ npairs(n) = loca(r,q,z); r=0;
forvec(p=vector(n,i,[1,2*n]),
q = eval( setminus( Set(vector(2*n,i,i)), Set(p) ) );
for(j=1,n!,
z=numtoperm(n,j);
if(type( sum(j=1,#p,p[j]/q[z[j]]) )=="t_INT",r++);
);
,2);
r
}
Regards,
Max
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