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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + n*b(n-1), where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
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%I #4 Dec 14 2017 14:23:01

%S 2,3,13,31,68,134,250,447,777,1323,2220,3697,6097,10002,16337,26609,

%T 43250,70199,113827,184444,298731,483679,782960,1267237,2050845,

%U 3318782,5370381,8689973,14061250,22752180,36814450,59567715,96383317,155952253,252336862

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + n*b(n-1), where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, 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="/A296291/b296291.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) = 1, a(1) = 4, b(0) = 2, b(1) = 3, b(2) = 5

%e a(2) = a(0) + a(1) + 2*b(1) = 11

%e Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, ...)

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

%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}]; (* A296291 *)

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

%Y Cf. A001622, A296245.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Dec 14 2017