/* This algorithm is taken from http://www.numericana.com/answer/numbers.htm#partitions
 * e=1 prints p[1],...,p[m] and returns p[m] if m>=1.
 * e=0 merely returns p[m].
 */
define partition(m,e){
auto i,p[],s,j,k,t,kk
    p[0] = 1

    for(i=1;i<=m;i++){
      j=1;k=1;s=0;
      while(j>0){
        kk=k*k
        j=i-(3*kk+k)/2
        t=parity(k)
        if(j>=0){
           s=s-t*p[j]
        }
        j=i-(3*kk-k)/2
        if(j>=0){
           s=s-t*p[j]
        }
        k=k+1
      }
      p[i]= s
      if(e){
         print "p[",i,"]=",s,"\n"
      }
    }
    return(s)
}

define parity(m){
if(m%2){
  return(-1)
}else{
  return(1)
}
}