A295962 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) - 1, where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
3, 4, 11, 20, 37, 64, 109, 182, 302, 496, 811, 1321, 2147, 3484, 5648, 9150, 14818, 23989, 38829, 62841, 101694, 164560, 266280, 430867, 697175, 1128071, 1825276, 2953378, 4778686, 7732097, 12510817, 20242949, 32753803, 52996790, 85750632, 138747462
Offset: 0
Examples
a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5 b(3) = 6 (least "new number") a(2) = a(1) + a(0) + b(2) - 1 = 11 Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
Links
- Clark Kimberling, Table of n, a(n) for n = 0..2000
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Programs
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Mathematica
a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; b[2] = 5; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n] - 1; j = 1; While[j < 6, k = a[j] - j - 1; While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; Table[a[n], {n, 0, k}]; (* A295962 *)
Comments