A295139 Solution of the complementary equation a(n) = 3*a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 6, 10, 23, 37, 77, 120, 242, 372, 739, 1130, 2232, 3406, 6713, 10236, 20158, 30728, 60495, 92206, 181509, 276643, 544553, 829956, 1633687, 2489897, 4901091, 7469722, 14703305, 22409199, 44109949, 67227632, 132329883
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4 a(2) =3*a(0) + b(0) = 6 Complement: (b(n)) = (3, 4, 5, 7, 8, 9, 11, 12, 13, 14, 15, ...)
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] = 2; b[0] = 3; b[1]=4; a[n_] := a[n] = 3 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}] (* A295139 *) Table[b[n], {n, 0, 10}]
Comments