A296245 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, 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, 28, 66, 143, 273, 497, 870, 1488, 2502, 4159, 6857, 11241, 18354, 29884, 48562, 78807, 127769, 207017, 335270, 542816, 878662, 1422103, 2301441, 3724273, 6026555, 9751728, 15779244, 25531996, 41312329, 66845481, 108159035, 175005812, 283166216
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5; a(2) = a(0) + a(1) + b(2)^2 = 28 Complement: (b(n)) = (3, 4, 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] = 2; b[0] = 3; b[1] = 4; b[2] = 5; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^2; j = 1; While[j < 12, 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}] (* A296245 *) 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