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A296292 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + n*b(n-1), where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, and (a(n)) and (b(n)) are increasing complementary sequences. 4

%I #4 Dec 14 2017 14:23:10

%S 2,4,12,31,67,133,248,444,772,1315,2217,3686,6083,9977,16298,26545,

%T 43147,70032,113557,184007,298024,482535,781109,1264242,2045999,

%U 3310941,5357694,8669445,14028035,22698437,36727492,59427014,96155658,155583893,251740843

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + n*b(n-1), where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, 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="/A296292/b296292.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.

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

%e a(2) = a(0) + a(1) + 2*b(1) = 12

%e Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, ...)

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

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

%t j = 1; While[j < 10, 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}]; (* A296292 *)

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

%Y Cf. A001622, A296245.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Dec 14 2017

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