A295140 Solution of the complementary equation a(n) = 3*a(n-2) - b(n-2) + 4, 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, 5, 9, 13, 24, 35, 66, 98, 190, 284, 559, 840, 1664, 2506, 4977, 7502, 14914, 22488, 44723, 67443, 134147, 202306, 402417, 606893, 1207225, 1820652, 3621647, 5461927, 10864911, 16385749, 32594700, 49157213, 97784065, 147471603, 293352158
Offset: 0
Examples
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4 a(2) =3*a(0) - b(0) + 4 = 5 Complement: (b(n)) = (2, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, ...)
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] = 3 a[n - 2] + b[n - 2] + 4; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; Table[a[n], {n, 0, 18}] (* A295140 *) Table[b[n], {n, 0, 10}]
Comments