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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
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%I #7 Nov 01 2017 23:04:51

%S 1,3,9,19,37,67,117,200,335,555,912,1491,2429,3948,6407,10387,16829,

%T 27253,44121,71415,115579,187039,302665,489753,792469,1282275,2074799,

%U 3357131,5431989,8789181,14221233,23010479,37231779,60242328,97474179,157716581

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

%C The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:

%C Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio. See A293358 for a guide to related sequences.

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

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

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

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

%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

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

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

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t Table[a[n], {n, 0, 40}] (* A294367 *)

%t Table[b[n], {n, 0, 10}]

%Y Cf. A001622 (golden ratio), A293765.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 29 2017