A296251 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^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, 19, 46, 101, 196, 361, 638, 1099, 1858, 3101, 5128, 8425, 13778, 22459, 36526, 59309, 96235, 155985, 252704, 409218, 662498, 1072341, 1735515, 2808585, 4544884, 7354310, 11900094, 19255365, 31156483, 50412937, 81570576, 131984738, 213556610, 345542717
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4; a(2) = a(0) + a(1) + b(1)^2 = 19; Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, ...)
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] = 2; b[0] = 3; b[1] = 4; 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}] (* A296251 *) 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