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A154811
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a(n) = Fibonacci(2n+1) mod 9.
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9
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1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1, 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1
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OFFSET
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0,2
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COMMENTS
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Periodic with period length 12. The period 1, 2, 5, 4, 7, 8, 8, 7, 4, 5, 2, 1 is the palindrome of the one in A153990.
There is some connection to A014217 using the formula with phi=(1+sqrt(5))/2 in A001519.
Terms of the simple continued fraction of 4343150/(sqrt(21657897254981)-1671809). [From Paolo P. Lava, Feb 17 2009]
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LINKS
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Table of n, a(n) for n=0..71.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,-1,1).
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FORMULA
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a(n) = A001519(n+1) mod 9 = A122367(n) mod 9 = |A099496(n)| mod 9.
a(n) = (1/132)*(9*(n mod 12) + 20*((n+1) mod 12) + 42*((n+2) mod 12) - 2*((n+3) mod 12) + 42*((n+4) mod 12) + 20*((n+5) mod 12) + 9*((n+6) mod 12) - 2*((n+7) mod 12) - 24*((n+8) mod 12) + 20*((n+9) mod 12) - 24*((n+10) mod 12) - 2*((n+11) mod 12)), with n >= 0. - Paolo P. Lava, Jan 16 2009
From R. J. Mathar, Apr 10 2009: (Start)
a(n) = a(n-1) - a(n-6) + a(n-7).
G.f.: -(1 + x + 3*x^2 - x^3 + 3*x^4 + x^5 + x^6)/((x - 1)*(x^2 + 1)*(x^4 - x^2 + 1)). (End)
a(n) = (270 - (14*m^5 - 175*m^4 + 760*m^3 - 1325*m^2 + 666*m + 210)*(-1)^floor(n/6))/60 where m = n - 6*floor(n/6). - Luce ETIENNE, Sep 17 2017
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MATHEMATICA
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Mod[Fibonacci[Range[1, 151, 2]], 9] (* Harvey P. Dale, Jul 10 2018 *)
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PROG
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(PARI) a(n)=fibonacci(n%12*2+1)%9 \\ Charles R Greathouse IV, Dec 21 2011
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CROSSREFS
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Cf. A010689, A010875.
Sequence in context: A085801 A023843 A153990 * A309513 A296203 A036237
Adjacent sequences: A154808 A154809 A154810 * A154812 A154813 A154814
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KEYWORD
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nonn,easy
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AUTHOR
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Paul Curtz, Jan 15 2009
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EXTENSIONS
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Edited by R. J. Mathar, Jan 23 2009
Typo in A-number in first formula corrected by R. J. Mathar, Feb 23 2009
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STATUS
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approved
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