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A296250 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, 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 A296250 #9 Dec 12 2017 08:41:07
%S A296250 3,4,32,72,153,289,523,912,1556,2612,4337,7145,11707,19108,31104,
%T A296250 50536,82001,132937,215379,348800,564708,914084,1479417,2394177,
%U A296250 3874323,6269284,10144448,16414632,26560041,42975762,69536959,112513946,182052201,294567516
%N A296250 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, 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 A296250 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 A296250 Clark Kimberling, <a href="/A296250/b296250.txt">Table of n, a(n) for n = 0..1000</a>
%H A296250 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 A296250 a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(2)^2 + f(n-2)*b(3)^2 + ... + f(2)*b(n-1)^2 + f(1)*b(n)^2, where f(n) = A000045(n), the n-th Fibonacci number.
%e A296250 a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5;
%e A296250 a(2) = a(0) + a(1) + b(2)^2 = 32;
%e A296250 Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...)
%t A296250 a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; b[2] = 5;
%t A296250 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^2;
%t A296250 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296250 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296250 Table[a[n], {n, 0, k}]   (* A296250 *)
%t A296250 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296250 Cf. A001622, A296245.
%K A296250 nonn,easy
%O A296250 0,1
%A A296250 _Clark Kimberling_, Dec 10 2017