A295949 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 8, 16, 31, 56, 97, 164, 273, 450, 737, 1202, 1956, 3176, 5151, 8347, 13519, 21888, 35430, 57342, 92797, 150165, 242989, 393182, 636200, 1029412, 1665644, 2695089, 4360767, 7055891, 11416694, 18472622, 29889354, 48362015, 78251409, 126613465, 204864916
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5; b(3) = 6 (least "new number"); a(2) = a(1) + a(0) + b(2) = 8; Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, ...)
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] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]; j = 1; While[j < 5, 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}] (* A295949 *) Table[b[n], {n, 0, 20}] (* complement *)
Formula
a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.
Comments