A295951 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), 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, 10, 19, 36, 63, 108, 182, 302, 497, 813, 1325, 2154, 3496, 5668, 9184, 14873, 24079, 38975, 63078, 102078, 165182, 267287, 432497, 699813, 1132340, 1832184, 2964556, 4796773, 7761363, 12558171, 20319571, 32877780, 53197390, 86075210, 139272641
Offset: 0
Examples
a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5; b(3) = 6 (least "new number"); a(2) = a(1) + a(0) + b(2) = 10; Complement: (b(n)) = (1, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, ...)
Links
- Clark Kimberling, Table of n, a(n) for n = 0..2000
- 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]; j = 1; While[j < 5, 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}] (* A295951 *) 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) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.
Comments