A295357 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 3, 5, 20, 42, 83, 149, 259, 438, 730, 1204, 1973, 3219, 5237, 8504, 13792, 22350, 36200, 58612, 94878, 153559, 248509, 402143, 650730, 1052954, 1703768, 2756809, 4460667, 7217569, 11678332, 18896000, 30574434, 49470539
Offset: 0
Examples
a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that b(3) = 7 (least "new number") a(3) = a(1) + a(0) + b(2) + b(1) + b(0) = 20 Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
Links
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Programs
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Mathematica
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + b[n - 3]; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; z = 32; u = Table[a[n], {n, 0, z}] (* A295357 *) v = Table[b[n], {n, 0, 10}] (* complement *)
Comments