A295138 Solution of the complementary equation a(n) = 3*a(n-2) + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 7, 11, 27, 41, 90, 133, 282, 412, 860, 1251, 2596, 3770, 7806, 11329, 23438, 34008, 70336, 102047, 211032, 306166, 633122, 918526, 1899395, 2755608, 5698216, 8266856, 17094681, 24800602, 51284078, 74401842
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3 a(2) =3*a(0) + b(1) = 7 Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ... )
Links
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Crossrefs
Cf. A295053.
Programs
-
Mathematica
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a[0] = 1; a[1] = 2; b[0] = 3; a[n_] := a[n] = 3 a[n - 2] + b[n - 1]; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; Table[a[n], {n, 0, 18}] (* A295138 *) Table[b[n], {n, 0, 10}]
Comments