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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. 3

%I

%S 1,3,29,68,146,278,505,883,1509,2536,4214,6946,11385,18587,30261,

%T 49172,79794,129366,209601,339451,549581,889608,1439814,2330098,

%U 3770641,6101523,9873064,15975548,25849636,41826273,67677065,109504563,177182924,286688856

%N 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 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 Clark Kimberling, <a href="/A296246/b296246.txt">Table of n, a(n) for n = 0..1000</a>

%H 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 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 a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5;

%e a(2) = a(0) + a(1) + b(2) = 29.

%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...).

%t a = 1; a = 3; b = 2; b = 4; b = 5;

%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n]^2;

%t j = 1; While[j < 6 , k = a[j] - j - 1;

%t While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];

%t Table[a[n], {n, 0, k}] (* A296246 *)

%t Table[b[n], {n, 0, 20}] (* complement *)

%Y Cf. A001622, A296245.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Feb 06 2018

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Last modified October 13 19:48 EDT 2019. Contains 327981 sequences. (Running on oeis4.)