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%I #12 Dec 12 2017 08:14:12
%S 2,3,6,25,56,130,250,461,811,1393,2348,3910,6454,10589,17299,28177,
%T 45800,74338,120538,195317,316339,512185,829100,1341961,2171790,
%U 3514535,5687166,9202601,14890728,24094353,38986170,63081679,102069074,165152049,267222492
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 2, a(1) = 3, b(0) = 1, 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.
%H Clark Kimberling, <a href="/A296259/b296259.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.
%F a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(0)^2 + f(n-2)*b(1)^2 + ... + f(2)*b(n-3)^2 + f(1)*b(n-2)^2, where f(n) = A000045(n), the n-th Fibonacci number.
%e a(0) = 2, a(1) = 3, b(0) = 1;
%e a(2) = a(0) + a(1) + b(0)^2 = 6;
%e Complement: (b(n)) = (1, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...)
%t a[0] = 2; a[1] = 3; b[0] = 1;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2]^2;
%t j = 1; While[j < 6 , 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}] (* A296259 *)
%t Table[b[n], {n, 0, 20}] (* complement *)
%Y Cf. A001622, A296245.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Dec 11 2017
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