A296253 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 4, 14, 43, 93, 185, 342, 608, 1050, 1779, 2973, 4921, 8119, 13296, 21704, 35324, 57389, 93113, 150943, 244540, 396012, 641128, 1037765, 1679569, 2718063, 4398416, 7117320, 11516636, 18634917, 30152577, 48788583, 78942316, 127732124, 206675736, 334409229
Offset: 0
Examples
a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3; a(2) = a(0) + a(1) + b(1)^2 = 14; Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, ...)
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; 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}] (* A296253 *) 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