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A287765
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Period 4: repeat [1, 3, 5, 3].
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1
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1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1
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OFFSET
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1,2
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LINKS
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FORMULA
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G.f.: x * (3*x^2+2*x+1) / (1-x+x^2-x^3). [Corrected by Georg Fischer, May 19 2019]
a(n) = a(n-1) - a(n-2) + a(n-3) with a(1)=1, a(2)=3 and a(3)=5.
a(2n) = 3, a(4*n+1) = 1 and a(4*n+3) = 5.
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MATHEMATICA
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PadRight[{}, 105, {1, 3, 5, 3}] (* or *)
f[n_] := Switch[Mod[n, 4], 0, 3, 1, 1, 2, 3, 3, 5]; Array[f, 105] (* or *)
CoefficientList[ Series[(3x^2 +2x +1)/(-x^3 +x^2 -x +1), {x, 0, 104}], x] (* or *)
LinearRecurrence[{1, -1, 1}, {1, 3, 5}, 105] (* or *)
RecurrenceTable[{a[n] == a[n - 1] - a[n - 2] + a[n - 3], a[1] == 1, a[2] == 3, a[3] == 5}, a, {n, 105}]
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CROSSREFS
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Inspired by the first difference of A108752.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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