A296255 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 4, 15, 44, 95, 188, 347, 616, 1063, 1800, 3007, 4976, 8179, 13411, 21879, 35614, 57854, 93868, 152163, 246515, 399207, 646298, 1046130, 1693104, 2739963, 4433851, 7174655, 11609406, 18785022, 30395452, 49181563, 79578171, 128760959, 208340426, 337102754
Offset: 0
Examples
a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3; a(2) = a(0) + a(1) + b(1)^2 = 15; Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
Programs
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Mathematica
a[0] = 2; a[1] = 4; b[0] = 1; b[1] = 3; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2; 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}] (* A296255 *) 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(1)^2 + f(n-2)*b(2)^2 + ... + f(2)*b(n-2)^2 + f(1)*b(n-1)^2, where f(n) = A000045(n), the n-th Fibonacci number.
Comments