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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) - 1, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
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%I #8 Feb 22 2018 22:43:10

%S 2,4,10,19,35,61,104,175,290,477,780,1271,2066,3353,5436,8808,14264,

%T 23093,37379,60495,97898,158418,256342,414787,671157,1085973,1757160,

%U 2843164,4600356,7443553,12043944,19487533,31531514,51019085,82550638,133569763

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) - 1, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, 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).

%C See A295862 for a guide to related sequences.

%H Clark Kimberling, <a href="/A295961/b295961.txt">Table of n, a(n) for n = 0..2000</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.

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

%e b(3) = 6 (least "new number")

%e a(2) = a(1) + a(0) + b(2) - 1 = 10

%e Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 20, ...)

%t a[0] = 2; a[1] = 4; b[0] = 1; b[1] = 3; b[2] = 5;

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

%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}]; (* A295961 *)

%Y Cf. A001622, A000045, A295862.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Dec 08 2017