A228470 a(n) = 6*a(n-2) + a(n-4), where a(0) = 3, a(1) = 11, a(2) = 18, a(3) = 68.

3, 11, 18, 68, 111, 419, 684, 2582, 4215, 15911, 25974, 98048, 160059, 604199, 986328, 3723242, 6078027, 22943651, 37454490, 141385148, 230804967, 871254539, 1422284292, 5368912382, 8764510719, 33084728831, 54009348606, 203877285368, 332820602355

Offset: 0

Author

Clark Kimberling, Aug 22 2013

Keywords

• nonn

Comments

Let d = A228469. Then a(n) is the least k > d(n) such that trace(k/d(n)) consists of the first n terms of 0101010101010101... See A228469.

Examples

			See A228469.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,6,0,1).

Crossrefs

Cf. A228469.

Programs

Mathematica

c1 = CoefficientList[Series[(2 + 8 x + x^2 + x^3)/(1 - 6 x^2 - x^4), {x, 0, 40}], x]; c2 = CoefficientList[Series[(3 + 11 x + 2 x^3)/(1 - 6 x^2 - x^4), {x, 0, 40}], x]; pairs = Transpose[CoefficientList[Series[{-((3 + 11 x + 2 x^3)/(-1 + 6 x^2 + x^4)), -((2 + 8 x + x^2 + x^3)/(-1 + 6 x^2 + x^4))}, {x, 0, 20}], x]]; t[{x_, y_, }] := t[{x, y}]; t[{x, y_}] := Prepend[If[# > y - #, {y - #, 1}, {#, 0}], y] &[Mod[x, y]]; userIn2[{x_, y_}] := Most[NestWhileList[t, {x, y}, (#[[2]] > 0) &]]; Map[Map[#[[3]] &, Rest[userIn2[#]]] &, pairs] (* Peter J. C. Moses, Aug 20 2013 *)
LinearRecurrence[{0, 6, 0, 1}, {3, 11, 18, 68}, 30] (* T. D. Noe, Aug 23 2013 *)