A296257 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 2, 12, 30, 67, 133, 249, 446, 776, 1322, 2219, 3710, 6125, 10060, 16441, 26790, 43555, 70706, 114661, 185808, 300953, 487290, 788819, 1276734, 2066229, 3343692, 5410705, 8755238, 14166904, 22923166, 37091159, 60015481, 97107865, 157124642, 254233876
Offset: 0
Examples
a(0) = 1, a(1) = 2, b(0) = 3; a(2) = a(0) + a(1) + b(0)^2 = 12; Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, ...)
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; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2]^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}] (* A296257 *) 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(0)^2 + f(n-2)*b(1)^2 + ... + f(2)*b(n-3)^2 + f(1)*b(n-2)^2, where f(n) = A000045(n), the n-th Fibonacci number.
Comments