A295947 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
2, 4, 11, 21, 39, 68, 116, 194, 322, 529, 865, 1409, 2290, 3716, 6024, 9759, 15803, 25584, 41410, 67018, 108453, 175497, 283977, 459502, 743508, 1203040, 1946579, 3149651, 5096263, 8245948, 13342246, 21588230, 34930513, 56518781, 91449334, 147968156
Offset: 0
Examples
a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, so that b(3) = 6 (least "new number"); a(2) = a(1) + a(0) + b(2) = 11; Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14, ...)
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] = 3; b[0] = 2; b[2] = 5; a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]; j = 1; While[j < 16, 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}]; (* A295947 *) 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