OFFSET
0,2
COMMENTS
The conditions that (a(n)) and (b(n)) be increasing and complementary force the equation a(n) = b(n-1) + b(a(n-1)), with initial value a(0) = 1, to have a unique solution; that is, as pair of complementary sequences (a(n)) = (1,5,11,10,...) and (b(n)) = (2,3,4,6,7,8,9,10,12,...). It appears that the 2nd difference sequence, (2,2,3,2,2,2,2,3,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,3,....), consists exclusively of 2s and 3s.
EXAMPLE
Start with a(1) = b(0) + b(a(0)) = b(0) + b(1) > = 2 + 3, so that b(0) = 2, b(1) = 3, b(2) = 4, and a(1) = 5. Then a(2) = b(1) + b(5) >= 11, so that b(5) = 6, etc.
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {};
Do[b = AppendTo[b, mex[Flatten[{a, b}], 1]], {2}];
a = AppendTo[a, b[[Length[a]]] + Last[b]];
Do[Do[b = AppendTo[b, mex[Flatten[{a, b}], 1]], {a[[-1]] - a[[-2]]}];
a = AppendTo[a, b[[Length[a]]] + Last[b]], {20}]; a
(* Peter J. C. Moses, Jul 14 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Aug 04 2019
STATUS
approved