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A228472 a(n) = 6*a(n-2) + a(n-4), where a(0) = 5, a(1) = 8, a(2) = 30, a(3) = 49. 1
5, 8, 30, 49, 185, 302, 1140, 1861, 7025, 11468, 43290, 70669, 266765, 435482, 1643880, 2683561, 10130045, 16536848, 62424150, 101904649, 384674945, 627964742, 2370473820, 3869693101, 14607517865, 23846123348, 90015581010, 146946433189, 554701003925 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
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.
LINKS
EXAMPLE
See A228471.
MATHEMATICA
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 *)
LinearRecurrence[{0, 6, 0, 1}, {5, 8, 30, 49}, 30] (* T. D. Noe, Aug 23 2013 *)
CROSSREFS
Sequence in context: A076593 A219775 A129774 * A361578 A049373 A304647
KEYWORD
nonn
AUTHOR
Clark Kimberling, Aug 22 2013
STATUS
approved

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Last modified May 21 17:21 EDT 2024. Contains 372738 sequences. (Running on oeis4.)