# A224762 and A224763

# Compute fractional curling number of an array or list a of length n.
fcn:=proc(a,n) local sw1,sw2,swt1,i1,i2,tmax,k,kp,lpp,lp,n1,n2,n3,t1;
if n=1 then RETURN(1); fi;
if n=2 then if a[1]=a[2] then RETURN(2) else RETURN(1); fi; fi;
k:=1;
# guess value of n1
for n1 from n by -1 to 2 do  # do n1
sw1:=-1;
# guess value of n2
for n2 from n1-1 by -1 to 1 do  # do n2
#lprint("n1,n2 ",n1,n2);
# do we have a match for P prime?
lpp:= n-n1+1; sw2:=-1;
      if lpp <= n2 then                    # if lpp
  for i1 from 1 to lpp do
  if a[n-i1+1] <> a[n2-i1+1] then sw2:=1; break; fi;
                       od:
#lprint("n,n1,n2,sw2,i1 ",n,n1,n2,sw2,i1);
if sw2 = -1 then                                            # if sw2
# is there an earlier copy of P?
n3:=n2-lpp+1; lp:=n1-n3;
#lprint("Have a copy of Pprime. n1,n2,lpp,lp ",n1,n2,lpp,lp);
# get max normal curling number of sequence without P prime
tmax:=floor((n1-1)/lp);
  for t1 from 0 to tmax-1 do     # do t1 
            if (n1-(t1+2)*lp < 1) then break; fi;                # if END
    swt1:=-1;
    for i2 from 1 to lp do         # do i2
      if a[n1-i2] <> a[n1-i2-(t1+1)*lp] then swt1:=1; break; fi;
                        od:        # od i2
    if swt1 = 1 then break; fi;
                        od:    # od t1
# now have a value of k
kp:=((t1+1)*lp+lpp)/lp;
#lprint("n,n1,n2,t1,kp ",n,n1,n2,t1,kp);
k:=max(k,kp);
fi;                                                        # fi sw2
        fi;                                 # fi lpp
                            od: # od n2
                         od: # od n1
RETURN(k); end proc;

# to compute M terms of A224762/A224763
g:=proc(M) local a,n;
a:=Array(1..M); a[1]:=1;
for n from 2 to M do a[n]:=fcn(a,n-1); od:
RETURN(a);
end proc;

g(61000);