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%I #11 Feb 22 2018 22:32:06
%S 2,4,11,21,39,68,116,194,322,529,865,1409,2290,3716,6024,9759,15803,
%T 25584,41410,67018,108453,175497,283977,459502,743508,1203040,1946579,
%U 3149651,5096263,8245948,13342246,21588230,34930513,56518781,91449334,147968156
%N 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.
%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
%C See A295862 for a guide to related sequences.
%H Clark Kimberling, <a href="/A295947/b295947.txt">Table of n, a(n) for n = 0..2000</a>
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%F 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.
%e a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, so that
%e b(3) = 6 (least "new number");
%e a(2) = a(1) + a(0) + b(2) = 11;
%e Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14, ...)
%t a[0] = 1; a[1] = 3; b[0] = 2; b[2] = 5;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
%t j = 1; While[j < 16, k = a[j] - j - 1;
%t While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t Table[a[n], {n, 0, k}]; (* A295947 *)
%t Table[b[n], {n, 0, 20}] (* complement *)
%Y Cf. A001622, A000045, A295862.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Dec 08 2017
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