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a(n) = 6*a(n-2) + a(n-4), where a(0) = 5, a(1) = 8, a(2) = 30, a(3) = 49.
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%I #21 Dec 16 2023 16:59:25

%S 5,8,30,49,185,302,1140,1861,7025,11468,43290,70669,266765,435482,

%T 1643880,2683561,10130045,16536848,62424150,101904649,384674945,

%U 627964742,2370473820,3869693101,14607517865,23846123348,90015581010,146946433189,554701003925

%N a(n) = 6*a(n-2) + a(n-4), where a(0) = 5, a(1) = 8, a(2) = 30, a(3) = 49.

%C Let d = A228471. Then a(n) is the least k > d(n) such that trace(k/d(n)) consists of the first n terms of 10101010101010101... See A228470.

%H Clark Kimberling, <a href="/A228472/b228472.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,1).

%e See A228471.

%t c1 = CoefficientList[Series[(3 + 5 x + x^2 + x^3)/(1 - 6 x^2 - x^4), {x, 0, 40}], x]; c2 = CoefficientList[Series[(5 + 8 x + 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 *)

%t LinearRecurrence[{0, 6, 0, 1}, {5, 8, 30, 49}, 30] (* _T. D. Noe_, Aug 23 2013 *)

%Y Cf. A228471, A228469.

%K nonn

%O 0,1

%A _Clark Kimberling_, Aug 22 2013