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A216676
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Digital roots of squares of Fibonacci numbers.
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5
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1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4
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OFFSET
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1,3
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COMMENTS
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The first 11 terms are symmetric about 6th term. The first 23 terms are symmetric about 12th term. We can generalize this as follows: the first (2n-1) terms are symmetric about n-th term.
The sequence appears to be periodic with period-length 12. - John W. Layman, Sep 14 2012
The Fibonacci numbers are periodic modulo any integer. The digital roots of the Fibonacci numbers are given by A030132, a sequence with a period length of 24. Squaring gives {1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9, 1, 1, 4, 9, 7, 1, 7, 9, 4, 1, 1, 9}, which is a sequence of twelve numbers given twice. Therefore, the previous comment is correct. - Alonso del Arte, Sep 25 2012
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LINKS
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FORMULA
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G.f. ( -1-x-3*x^2-8*x^3-3*x^4+8*x^5-9*x^7-x^6 ) / ( (x-1) *(1+x) *(x^2+1) *(x^4-x^2+1) ). - R. J. Mathar, Sep 15 2012
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EXAMPLE
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a(7) = 7 because F(7) = 13 and 13^2 = 169, with digits adding up to 16, the digital root is therefore 7.
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MATHEMATICA
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a = {}; For[n = 1, n <= 100, n++, {fn2 = Fibonacci[n]^2; d = IntegerDigits[fn2]; While[Length[d] > 1, d = IntegerDigits[Total[d]]]; AppendTo[a, d[[1]]] }]; a (* John W. Layman, Sep 14 2012 *)
ReplaceAll[Table[Mod[Fibonacci[n]^2, 9], {n, 72}], {0 -> 9}] (* Alonso del Arte, Sep 23 2012 *)
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PROG
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(PARI) fibmod(n, m)=((Mod([1, 1; 1, 0], m))^n)[1, 2]
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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