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A295962 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) - 1, 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 A295962 #8 Feb 22 2018 22:43:24
%S A295962 3,4,11,20,37,64,109,182,302,496,811,1321,2147,3484,5648,9150,14818,
%T A295962 23989,38829,62841,101694,164560,266280,430867,697175,1128071,1825276,
%U A295962 2953378,4778686,7732097,12510817,20242949,32753803,52996790,85750632,138747462
%N A295962 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) - 1, 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 A295962 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 A295962 See A295862 for a guide to related sequences.
%H A295962 Clark Kimberling, <a href="/A295962/b295962.txt">Table of n, a(n) for n = 0..2000</a>
%H A295962 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e A295962 a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5
%e A295962 b(3) = 6 (least "new number")
%e A295962 a(2) = a(1) + a(0) + b(2) - 1 = 11
%e A295962 Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
%t A295962 a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; b[2] = 5;
%t A295962 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n] - 1;
%t A295962 j = 1; While[j < 6, k = a[j] - j - 1;
%t A295962 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A295962 Table[a[n], {n, 0, k}];  (* A295962 *)
%Y A295962 Cf. A001622, A000045, A295862.
%K A295962 nonn,easy
%O A295962 0,1
%A A295962 _Clark Kimberling_, Dec 08 2017