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A296252 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296252 #10 Dec 12 2017 08:41:37
%S A296252 1,3,20,48,104,201,369,651,1120,1892,3156,5217,8569,14011,22836,37136,
%T A296252 60296,97793,158530,256807,415866,673249,1089740,1763665,2854134,
%U A296252 4618583,7473558,12093041,19567560,31661625,51230274,82893055,134124554,217018905,351144828
%N A296252 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296252 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 A296252 Clark Kimberling, <a href="/A296252/b296252.txt">Table of n, a(n) for n = 0..1000</a>
%H A296252 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 A296252 a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(1)^2 + f(n-2)*b(2)^2 + ... + f(2)*b(n-2)^2 + f(1)*b(n-1)^2, where f(n) = A000045(n), the n-th Fibonacci number.
%e A296252 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4;
%e A296252 a(2) = a(0) + a(1) + b(1)^2 = 20;
%e A296252 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, ...)
%t A296252 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
%t A296252 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;
%t A296252 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296252 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296252 Table[a[n], {n, 0, k}]   (* A296252 *)
%t A296252 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296252 Cf. A001622, A296245.
%K A296252 nonn,easy
%O A296252 0,2
%A A296252 _Clark Kimberling_, Dec 11 2017