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A295949 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295949 #10 Feb 22 2018 22:33:59
%S A295949 1,2,8,16,31,56,97,164,273,450,737,1202,1956,3176,5151,8347,13519,
%T A295949 21888,35430,57342,92797,150165,242989,393182,636200,1029412,1665644,
%U A295949 2695089,4360767,7055891,11416694,18472622,29889354,48362015,78251409,126613465,204864916
%N A295949 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295949 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 A295949 See A295862 for a guide to related sequences.
%H A295949 Clark Kimberling, <a href="/A295949/b295949.txt">Table of n, a(n) for n = 0..2000</a>
%H A295949 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 A295949 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 A295949 a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5;
%e A295949 b(3) = 6 (least "new number");
%e A295949 a(2) = a(1) + a(0) + b(2) = 8;
%e A295949 Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17,  ...)
%t A295949 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5;
%t A295949 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
%t A295949 j = 1; While[j < 5, k = a[j] - j - 1;
%t A295949 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A295949 Table[a[n], {n, 0, k}]  (* A295949 *)
%t A295949 Table[b[n], {n, 0, 20}] (* complement *)
%Y A295949 Cf. A001622, A000045, A295862.
%K A295949 nonn,easy
%O A295949 0,2
%A A295949 _Clark Kimberling_, Dec 08 2017