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A086695
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a(n) = 100^n - 10^n - 1.
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0
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OFFSET
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0,1
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COMMENTS
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Digits of the inverse of these numbers gives the Fibonacci numbers. More precisely the digits of 1/(10^(2*n)-10^n-1) give the Fibonacci numbers up to 10^n.
More generally, if x_1,x_2, x_n=x_(n-1)-x_(n-2) is any Lucas sequence, then the digits of the numbers (x_1*10^n-(x_1-x_2))/(10^(2*n)-10^n-1) gives the x_n up to 10^n.
1/a(n) = Sum(i=1..infinity) A000045(i-1)/10^(n*i) (see Long paper). - Michel Marcus, May 01 2013
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LINKS
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Table of n, a(n) for n=0..9.
C. T. Long, The Decimal Expansion of 1/89 and Related Results, The Fibonacci Quarterly, Volume 19, Number 1, February 1981
Index to sequences with linear recurrences with constant coefficients, signature (111,-1110,1000).
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FORMULA
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a(n) = 10^(2*n)-10^n-1.
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PROG
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(PARI) a(n)=100^n-10^n-1 \\ Charles R Greathouse IV, May 01 2013
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CROSSREFS
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Sequence in context: A020472 A093948 A116254 * A056568 A174758 A181681
Adjacent sequences: A086692 A086693 A086694 * A086696 A086697 A086698
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KEYWORD
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easy,nonn
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AUTHOR
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Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003
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STATUS
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approved
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