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Solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.
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%I #12 Jun 25 2018 17:36:52

%S 1,3,10,17,21,25,29,33,38,45,49,53,57,61,66,73,77,82,89,93,98,105,109,

%T 114,121,125,130,137,141,145,150,157,161,165,169,173,178,185,189,194,

%U 201,205,210,217,221,226,233,237,242,249,253,257,262,269,273,277

%N Solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295862 for a guide to related sequences.

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

%e a(2) = a(0)*b(1) + a(1)*b(0) = 10

%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 18, ...)

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

%t a[0] = 1; a[1] = 3; b[0] = 2;

%t a[n_] := a[n] = a[0]*b[n - 1] + a[1]*b[n - 2];

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t u = Table[a[n], {n, 0, 500}]; (* A296219 *)

%t Table[b[n], {n, 0, 20}]

%Y Cf. A296000.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Dec 08 2017

%E Conjectured g.f. removed by _Alois P. Heinz_, Jun 25 2018