A295062 Solution of the complementary equation a(n) = 4*a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 3, 6, 16, 29, 71, 124, 293, 506, 1183, 2036, 4745, 8158, 18995, 32649, 75998, 130615, 304012, 522481, 1216070, 2089947, 4864304, 8359813, 19457242, 33439279, 77828996, 133757146, 311316015
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4 a(2) = 4*a(0) + b(0) = 6 Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, ...)
Links
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Crossrefs
Cf. A295053.
Programs
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Mathematica
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a[0] = 1; a[1] = 3; b[0] = 2; b[1]=4; a[n_] := a[n] = 4 a[n - 2] + b[n - 2]; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; Table[a[n], {n, 0, 18}] (* A295062 *) Table[b[n], {n, 0, 10}]
Comments