proposed

approved

editing

proposed

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

1, 3, 8, 17, 38, 75, 161, 310, 655, 1252, 2633, 5022, 10547, 20104, 42206, 80435, 168844, 321761, 675398, 1287067, 2701616, 5148293, 10806490, 20593199, 43225988, 82372825, 172903982

0,2

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.

The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.09... and 2.09... .

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

a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

a(2) = 4*a(0) + b(1) = 8

Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ...)

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

a[0] = 1; a[1] = 3; b[0] = 2;

a[n_] := a[n] = 4 a[n - 2] + b[n - 1];

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

Table[a[n], {n, 0, 18}] (* A295061 *)

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

Cf. A295053.

allocated

nonn,easy

Clark Kimberling, Nov 18 2017

approved

editing

allocated for Clark Kimberling

allocated

approved