\datethis
@*Intro. Sequence M1805, posets with linear order $1\ldots n$.
Same as upper triangular $n\times n$ Boolean matrices $B$
with zeros on diagonal and $B^2\subseteq B$.

The unique(?) thing here is that I use $2^k$ as an index, not $k$;
therefore I can also use $2^k+2^l$ as an index into a triangular matrix!

@d maxn 9
@d maxnn (1<<(maxn-1))
@c
#include <stdio.h>
int row[maxnn+1]; /* |row[1<<(n-j)]| is $j$th row of $B$ */
int mask[maxnn+(maxnn>>1)+1];
  /* |mask[1<<(n-j)]| shows bits that must be zero in $j$th row */
int sols;

main()
{
  register int l,x,y,z;
  int n,nn;
  for (n=3;n<=maxn;n++) {
    sols=0;
    l=nn=1<<(n-1);
    for (x=2;x<=l;x<<=1) mask[x]=0;
 newlev:@+if (l==2) {
      sols+=2-(mask[2]&1);
      goto backtrack;
    }
    mask[l]&=l-1;
    row[l]=0;
    l>>=1; goto newlev;
 backtrack: l<<=1, x=row[l];
    for (y=x&(x+1);y;y-=z) z=y&-y, mask[z]=mask[l+z];
    x=(x|mask[l])+1;
    if (x>=l) {
      if (l==nn) goto done;
      goto backtrack;
    }
    row[l]=x=x&~mask[l];
    for (y=x&(x+1),x=x^-1;y;y-=z) z=y&-y, mask[l+z]=mask[z], mask[z]|=x;
    l>>=1; goto newlev;
 done: printf("%d solutions for %d.\n",sols,n);
  }
}

@*Index.