cp's OEIS Frontend

A296255 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296255 #10 Dec 12 2017 08:13:32
%S A296255 2,4,15,44,95,188,347,616,1063,1800,3007,4976,8179,13411,21879,35614,
%T A296255 57854,93868,152163,246515,399207,646298,1046130,1693104,2739963,
%U A296255 4433851,7174655,11609406,18785022,30395452,49181563,79578171,128760959,208340426,337102754
%N A296255 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296255 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 A296255 Clark Kimberling, <a href="/A296255/b296255.txt">Table of n, a(n) for n = 0..1000</a>
%H A296255 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 A296255 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 A296255 a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3;
%e A296255 a(2) = a(0) + a(1) + b(1)^2 = 15;
%e A296255 Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)
%t A296255 a[0] = 2; a[1] = 4; b[0] = 1; b[1] = 3;
%t A296255 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;
%t A296255 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296255 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296255 Table[a[n], {n, 0, k}]   (* A296255 *)
%t A296255 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296255 Cf. A001622, A296245.
%K A296255 nonn,easy
%O A296255 0,1
%A A296255 _Clark Kimberling_, Dec 11 2017