%I #9 Oct 15 2019 18:16:55
%S 1,1,2,65,37709445313,
%T 44237187584945316063023352626539792098172619389908174093304834
%N a(1)=a(2)=1, a(n+1) = (a(n)^6 +1)/a(n-1).
%p A003821 := proc(n) option remember; if n <= 2 then 1 else (A003821(n-1)^6+1)/A003821(n-2); fi; end;
%t RecurrenceTable[{a[1]==a[2]==1,a[n+1]==(a[n]^6+1)/a[n-1]},a,{n,6}] (* _Harvey P. Dale_, Oct 15 2019 *)
%K nonn
%O 1,3
%A Waldemar Pompe (pompe(AT)zodiac1.mimuw.edu.pl)