## calculate the value of sum[n=1..oo] x^(2^(-n))-1
## i.e. sum of repeated square roots, less one.

define ss(x) {
  auto s,t,os;
  if(x<=0){print "Negative Infinity\n";-1/0}
  scale+=6
  s=x ; os=s+1
  t=0;while(os!=s){
    os=s
    s=sqrt(s)
    t+=s-1
  }
  scale-=6;t/=1
  return(t);
}

define ss_n(n,x) { # Generalisation of ss; ss(x) == ss_n(2,x)
  auto s,t,os;
  if(x<=0){print "Negative Infinity\n";-1/0}
  if(n==2)return(ss(x))
  if(n<=1){print "Infinity\n";1/0}
  scale+=6
  s=x ; os=s+1
  t=0;while(os!=s){
    os=s
    s=e(l(s)/n)
    t+=s-1
  }
  scale-=6;t/=1
  return(t);
}