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A296280 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)*b(n-1)*b(n), where a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296280 #4 Dec 13 2017 18:39:58
%S A296280 1,4,35,129,374,839,1717,3276,5983,10579,18278,31041,52049,86450,
%T A296280 142579,233925,382318,623083,1013381,1645704,2669711,4327559,7011070,
%U A296280 11354229,18382849,29756734,48161507,77942601,126131078,204103439,330267253,534406596,864714241
%N A296280 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)*b(n-1)*b(n), where a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296280 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). See A296245 for a guide to related sequences.
%H A296280 Clark Kimberling, <a href="/A296280/b296280.txt">Table of n, a(n) for n = 0..1000</a>
%H A296280 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 A296280 a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5
%e A296280 a(2) = a(0) + a(1) + b(0)*b(1)*b(2) = 35
%e A296280 Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...)
%t A296280 a[0] = 1; a[1] = 4; b[0] = 2; b[1] = 3; b[2] = 5;
%t A296280 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2] b[n - 1] b[n];
%t A296280 j = 1; While[j < 10, k = a[j] - j - 1;
%t A296280  While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296280 Table[a[n], {n, 0, k}]; (* A296280 *)
%t A296280 Table[b[n], {n, 0, 20}]    (* complement *)
%Y A296280 Cf. A001622, A296245.
%K A296280 nonn,easy
%O A296280 0,2
%A A296280 _Clark Kimberling_, Dec 13 2017