# A305369 June 2 2018
read format;
with(Bits):

# set M = limits on arrays, F = number of terms needed
# a = sequence, q = LK = link to next term, p = link to previous term
# smn = smallest missing number
# have[i] = 1 iff i in sequence

M:=50000; F:=10000;
have:=Array(1..M,-1): a:=Array(1..M): LK:=Array(1..M): SMN:=Array(1..M):

# initiate sequence
a[1]:=1; smn:=2; p:=0; q:=1; have[1]:=1; LK[1]:=1; SMN[1]:=2;

# get n-th term
for n from 2 to F do
   for j from smn to M do 
   if ((have[j]=-1) and (And(j,q)=q) and (And(j,p)=0)) then 
      a[n]:=j; have[j]:=1; p:=q; q:=j-q;  LK[n]:=q;
# update smn?
	if j=smn then
for i from 1 to M do if have[i] <> 1 then smn:=i; break; fi; od:
	fi;
      SMN[n]:=smn;
      break; fi;
   od; # od j
od; # od n

# [seq(a[n],n=1..F)]; # A305369
# [seq(LK[n],n=1..F)]; # A305371