cp's OEIS Frontend

A295952 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 5, b(0) = 2, b(1) = 3, b(2) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.

%I A295952 #9 Feb 22 2018 22:40:19
%S A295952 1,5,10,21,38,67,114,192,318,523,855,1393,2264,3674,5956,9649,15625,
%T A295952 25296,40944,66264,107233,173523,280783,454334,735146,1189510,1924687,
%U A295952 3114229,5038949,8153212,13192196,21345444,34537677,55883160,90420877,146304078
%N A295952 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 5, b(0) = 2, b(1) = 3, b(2) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
%C A295952 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).
%C A295952 See A295862 for a guide to related sequences.
%H A295952 Clark Kimberling, <a href="/A295952/b295952.txt">Table of n, a(n) for n = 0..2000</a>
%H A295952 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 A295952 a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.
%e A295952 a(0) = 1, a(1) = 5, b(0) = 2, b(1) = 3, b(2) = 4;
%e A295952 b(3) = 6 (least "new number");
%e A295952 a(2) = a(1) + a(0) + b(2) = 10;
%e A295952 Complement: (b(n)) = (2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, ...)
%t A295952 a[0] = 1; a[1] = 5; b[0] = 2; b[1] = 3; b[2] = 4;
%t A295952 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
%t A295952 j = 1; While[j < 5, k = a[j] - j - 1;
%t A295952 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
%t A295952 Table[a[n], {n, 0, k}]   (* A295952 *)
%t A295952 Table[b[n], {n, 0, 20}]  (* complement *)
%Y A295952 Cf. A001622, A000045, A295862.
%K A295952 nonn,easy
%O A295952 0,2
%A A295952 _Clark Kimberling_, Dec 08 2017