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A295948 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295948 #10 Feb 22 2018 22:33:18
%S A295948 3,4,12,22,41,71,121,202,334,549,897,1461,2374,3852,6244,10115,16379,
%T A295948 26515,42917,69456,112398,181880,294305,476213,770547,1246790,2017368,
%U A295948 3264190,5281591,8545815,13827441,22373292,36200770,58574100,94774909,153349049
%N A295948 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295948 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 A295948 See A295862 for a guide to related sequences.
%H A295948 Clark Kimberling, <a href="/A295948/b295948.txt">Table of n, a(n) for n = 0..2000</a>
%H A295948 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 A295948 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 A295948 a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, so that
%e A295948 b(3) = 6 (least "new number");
%e A295948 a(2) = a(1) + a(0) + b(2) = 12;
%e A295948 Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)
%t A295948 a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; b[2] = 5;
%t A295948 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
%t A295948 j = 1; While[j < 16, k = a[j] - j - 1;
%t A295948 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A295948 Table[a[n], {n, 0, k}]  (* A295948 *)
%t A295948 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A295948 Cf. A001622, A000045, A295862.
%K A295948 nonn,easy
%O A295948 0,1
%A A295948 _Clark Kimberling_, Dec 08 2017