A296248 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 3, 30, 69, 148, 281, 510, 891, 1522, 2557, 4248, 7001, 11474, 18731, 30494, 49549, 80404, 130353, 211198, 342035, 553762, 896373, 1450760, 2347809, 3799298, 6147891, 9948030, 16096882, 26045936, 42143907, 68190999, 110336131, 178528426, 288865926
Offset: 0
Examples
a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5; a(2) = a(0) + a(1) + b(2) = 30 Complement: (b(n)) = (1, 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] = 2; a[1] = 3; b[0] = 1; b[1] = 4; 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}] (* A296248 *) 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