%I #5 Nov 01 2017 12:27:23

%S 1,3,9,14,23,31,43,55,70,85,103,122,143,165,189,214,241,269,299,331,

%T 364,399,435,473,512,553,596,640,686,733,782,832,884,937,992,1048,

%U 1106,1166,1227,1290,1354,1420,1487,1556,1626,1698,1771,1846,1923,2001,2081

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

%C The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.

%H Clark Kimberling, <a href="/A294480/b294480.txt">Table of n, a(n) for n = 0..1000</a>

%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, so that

%e a(2) = a(0) + b(1) = 9

%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 12, 13, 15, ...)

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

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

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

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

%t Table[a[n], {n, 0, 40}] (* A294480 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A293076, A293765, A294476.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Nov 01 2017