A295367 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)*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, 15, 37, 82, 161, 299, 532, 921, 1563, 2616, 4335, 7133, 11692, 19097, 31095, 50534, 82009, 132963, 215434, 348903, 564889, 914392, 1479931, 2395025, 3875712, 6271549, 10148131, 16420610, 26569733, 42991399, 69562254, 112554843
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that b(2) = 5 (least "new number") a(2) = a(1) + a(0) + b(1)*b(0) = 15 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, ...)
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] = 2; b[0] = 3; b[1] = 4; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1]*b[n - 2]; 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}] (* A295367 *) v = Table[b[n], {n, 0, 10}] (* complement *)
Formula
a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
Comments