cp's OEIS Frontend

A296256 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296256 #10 Dec 12 2017 08:13:48
%S A296256 3,4,11,40,87,176,327,584,1011,1739,2919,4854,7998,13108,21395,34827,
%T A296256 56583,91810,148834,241128,390491,632195,1023311,1656182,2680222,
%U A296256 4337188,7018251,11356339,18375551,29732914,48109554,77843624,125954403,203799323,329755095
%N A296256 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296256 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 A296256 Clark Kimberling, <a href="/A296256/b296256.txt">Table of n, a(n) for n = 0..1000</a>
%H A296256 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 A296256 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 A296256 a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2;
%e A296256 a(2) = a(0) + a(1) + b(1)^2 = 11;
%e A296256 Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 12, 13, 14, ...)
%t A296256 a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2;
%t A296256 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;
%t A296256 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296256 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296256 Table[a[n], {n, 0, k}]   (* A296256 *)
%t A296256 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296256 Cf. A001622, A296245.
%K A296256 nonn,easy
%O A296256 0,1
%A A296256 _Clark Kimberling_, Dec 11 2017