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A296246 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296246 #7 Aug 25 2019 13:42:11
%S A296246 1,3,29,68,146,278,505,883,1509,2536,4214,6946,11385,18587,30261,
%T A296246 49172,79794,129366,209601,339451,549581,889608,1439814,2330098,
%U A296246 3770641,6101523,9873064,15975548,25849636,41826273,67677065,109504563,177182924,286688856
%N A296246 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296246 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 A296246 Clark Kimberling, <a href="/A296246/b296246.txt">Table of n, a(n) for n = 0..1000</a>
%H A296246 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 A296246 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 A296246 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5;
%e A296246 a(2) = a(0) + a(1) + b(2) = 29.
%e A296246 Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...).
%t A296246 a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4; b[2] = 5;
%t A296246 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^2;
%t A296246 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296246 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296246 Table[a[n], {n, 0, k}]   (* A296246 *)
%t A296246 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296246 Cf. A001622, A296245.
%K A296246 nonn,easy
%O A296246 0,2
%A A296246 _Clark Kimberling_, Feb 06 2018