A296249 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 4, 31, 71, 151, 286, 518, 904, 1543, 2591, 4303, 7090, 11618, 18964, 30871, 50159, 81391, 131950, 213782, 346216, 560527, 907319, 1468471, 2376466, 3845666, 6222916, 10069423, 16293239, 26363686, 42658014, 69022856, 111682095, 180706247, 292389711
Offset: 0
Examples
a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5; a(2) = a(0) + a(1) + b(2)^2 = 31; Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...)
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; 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}] (* A296249 *) 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