A293350 - OEIS

%I #7 Nov 02 2017 09:19:44

%S 1,3,10,23,46,85,150,257,432,718,1182,1935,3155,5131,8330,13508,21888,

%T 35449,57393,92901,150356,243323,393748,637143,1030966,1668187,

%U 2699234,4367505,7066826,11434421,18501340,29935857,48437296,78373255,126810656,205184019

%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 2n, 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.  See A293076 for a guide to related sequences.

%C Conjecture:  a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.

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

%e a(2)  = a(1) + a(0) + b(0) + 4 = 10;

%e a(3) = a(2) + a(1) + b(1) + 6 = 23.

%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14,...)

%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 - 2] + 2n;

%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}]  (* A293350 *)

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

%Y Cf. A001622 (golden ratio), A293076.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 28 2017