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A130893
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Lucas numbers (beginning with 1) mod 10.
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8
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1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4
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OFFSET
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0,2
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COMMENTS
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Period 12: repeat [1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9, 2].
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
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FORMULA
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a(n) = (a(n-2) + a(n-1)) mod 10, with a(0) = 1, a(1) = 3.
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6) + a(n-7) - a(n-8) + a(n-9) - a(n-10) + a(n-11).
a(n+12) = a(n). (End)
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EXAMPLE
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1 + 3 = 4 = 4 mod 10, then a(3) = 4.
3 + 4 = 7 = 7 mod 10, then a(4) = 7.
4 + 7 = 11 = 1 mod 10, then a(5) = 1.
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MATHEMATICA
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Nest[Append[#, Mod[Total[Take[#, -2]], 10]] &, {1, 3}, 110] (* Harvey P. Dale, Apr 05 2011 *)
t = {1, 3}; Do[AppendTo[t, Mod[t[[-1]] + t[[-2]], 10]], {99}]; t (* T. D. Noe, Sep 16 2013 *)
LinearRecurrence[{1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1}, {1, 3, 4, 7,
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PROG
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(Ruby)
def truncM10(n)
..a = 1
..b = 3
..n.times do
....a, b = (b % 10), ((a + b) % 10)
..end
..return b
end
(PARI) a(n) = (fibonacci(n+1)+fibonacci(n-1)) % 10;
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 22 2007
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EXTENSIONS
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STATUS
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approved
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