A296247 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 4, 30, 70, 149, 283, 513, 896, 1530, 2570, 4269, 7035, 11529, 18820, 30638, 49782, 80781, 130963, 212185, 343632, 556346, 900554, 1457525, 2358755, 3817009, 6176548, 9994398, 16171907, 26167329, 42340325, 68508810, 110850360, 179360466, 290212195
Offset: 0
Examples
a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5; a(2) = a(0) + a(1) + b(2) = 30 Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 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] = 1; a[1] = 4; b[0] = 2; b[1] = 3; b[2] = 5; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^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}] (* A296247 *) 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)^2 + f(n-2)*b(3)^2 + ... + f(2)*b(n-1)^2 + f(1)*b(n)^2, where f(n) = A000045(n), the n-th Fibonacci number.
Comments