cp's OEIS Frontend

A296248 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A296248 #6 Dec 11 2017 14:31:34
%S A296248 2,3,30,69,148,281,510,891,1522,2557,4248,7001,11474,18731,30494,
%T A296248 49549,80404,130353,211198,342035,553762,896373,1450760,2347809,
%U A296248 3799298,6147891,9948030,16096882,26045936,42143907,68190999,110336131,178528426,288865926
%N A296248 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n)^2, where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A296248 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 A296248 Clark Kimberling, <a href="/A296248/b296248.txt">Table of n, a(n) for n = 0..1000</a>
%H A296248 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 A296248 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 A296248 a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, b(2) = 5;
%e A296248 a(2) = a(0) + a(1) + b(2) = 30
%e A296248 Complement: (b(n)) = (1, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...)
%t A296248 a[0] = 2; a[1] = 3; b[0] = 1; b[1] = 4; b[2] = 5;
%t A296248 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^2;
%t A296248 j = 1; While[j < 6 , k = a[j] - j - 1;
%t A296248 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A296248 Table[a[n], {n, 0, k}]     (* A296248 *)
%t A296248 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A296248 Cf. A001622, A296245.
%K A296248 nonn,easy
%O A296248 0,1
%A A296248 _Clark Kimberling_, Dec 10 2017