%I #6 Dec 12 2017 16:38:24
%S 1,4,20,54,116,226,414,730,1254,2116,3526,5824,9560,15624,25456,41386,
%T 67184,108969,176615,286090,463257,749947,1213854,1964503,3179113,
%U 5144428,8324411,13469769,21795172,35265997,57062291,92329478,149393029,241723839,391118274
%N Solution of the complementary equation a(n) = a(n-1) + a(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 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.
%C a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5
%C a(2) = a(0) + a(1) + b(1)*b(2) = 20
%C Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
%C a[0] = 1; a[1] = 4; b[0] = 2; b[1] = 3; b[2] = 5;
%C a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] b[n];
%C j = 1; While[j < 10, k = a[j] - j - 1;
%C While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%C Table[a[n], {n, 0, k}]; (* A296274 *)
%C Table[b[n], {n, 0, 20}] (* complement *)
%C Cf. A001622, A296245.
%C easy
%H Clark Kimberling, <a href="/A296274/b296274.txt">Table of n, a(n) for n = 0..1000</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.
%e a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5;
%e a(2) = a(0) + a(1) + b(1)*b(2) = 20;
%e Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
%t a[0] = 1; a[1] = 4; b[0] = 2; b[1] = 3; b[2] = 5;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] b[n];
%t j = 1; While[j < 10, 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}]; (* A296274 *)
%t Table[b[n], {n, 0, 20}] (* complement *)
%Y Cf. A001622, A296245.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Dec 12 2017