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A296251 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296251 #9 Dec 12 2017 08:41:16
%S A296251 1,2,19,46,101,196,361,638,1099,1858,3101,5128,8425,13778,22459,36526,
%T A296251 59309,96235,155985,252704,409218,662498,1072341,1735515,2808585,
%U A296251 4544884,7354310,11900094,19255365,31156483,50412937,81570576,131984738,213556610,345542717
%N A296251 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296251 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 A296251 Clark Kimberling, <a href="/A296251/b296251.txt">Table of n, a(n) for n = 0..1000</a>
%H A296251 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 A296251 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 A296251 a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4;
%e A296251 a(2) = a(0) + a(1) + b(1)^2 = 19;
%e A296251 Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, ...)
%t A296251 a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
%t A296251 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;
%t A296251 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296251 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296251 Table[a[n], {n, 0, k}]   (* A296251 *)
%t A296251 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296251 Cf. A001622, A296245.
%K A296251 nonn,easy
%O A296251 0,2
%A A296251 _Clark Kimberling_, Dec 10 2017