A326665 - OEIS

%I #6 Aug 07 2019 06:59:53

%S 1,5,11,19,30,43,58,75,94,116,140,166,194,224,256,290,327,366,407,450,

%T 495,542,591,642,695,750,808,868,930,994,1060,1128,1198,1270,1344,

%U 1420,1498,1578,1661,1746,1833,1922,2013,2106,2201,2298,2397,2498,2601,2706

%N Solution sequence (a(n)) of the complementary equation a(n) = b(n-1) + b(a(n-1)), for n >= 1, with a(0) = 1.

%C 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.

%e 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.

%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

%t a = {1}; b = {};

%t Do[b = AppendTo[b, mex[Flatten[{a, b}], 1]], {2}];

%t a = AppendTo[a, b[[Length[a]]] + Last[b]];

%t Do[Do[b = AppendTo[b, mex[Flatten[{a, b}], 1]], {a[[-1]] - a[[-2]]}];

%t a = AppendTo[a, b[[Length[a]]] + Last[b]], {20}]; a

%t (* _Peter J. C. Moses_, Jul 14 2019 *)

%Y Cf. A022424.

%K nonn

%O 0,2

%A _Clark Kimberling_, Aug 04 2019