A295952 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 5, b(0) = 2, b(1) = 3, b(2) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
1, 5, 10, 21, 38, 67, 114, 192, 318, 523, 855, 1393, 2264, 3674, 5956, 9649, 15625, 25296, 40944, 66264, 107233, 173523, 280783, 454334, 735146, 1189510, 1924687, 3114229, 5038949, 8153212, 13192196, 21345444, 34537677, 55883160, 90420877, 146304078
Offset: 0
Examples
a(0) = 1, a(1) = 5, b(0) = 2, b(1) = 3, b(2) = 4; b(3) = 6 (least "new number"); a(2) = a(1) + a(0) + b(2) = 10; Complement: (b(n)) = (2, 3, 4, 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] = 1; a[1] = 5; b[0] = 2; b[1] = 3; b[2] = 4; 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}] (* A295952 *) 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