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%I
%S 89,9899,998999,99989999,9999899999,999998999999,99999989999999,
%T 9999999899999999,999999998999999999,99999999989999999999
%N a(n) = 100^n - 10^n - 1.
%C 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.
%C 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.
%C 1/a(n) = Sum(i=1..infinity) A000045(i-1)/10^(n*i) (see Long paper). - _Michel Marcus_, May 01 2013
%H C. T. Long, <a href="http://www.fq.math.ca/Scanned/19-1/long.pdf">The Decimal Expansion of 1/89 and Related Results</a>, The Fibonacci Quarterly, Volume 19, Number 1, February 1981
%H <a href="/index/Rea#recLCC">Index to sequences with linear recurrences with constant coefficients</a>, signature (111,-1110,1000).
%F a(n) = 10^(2*n)-10^n-1.
%o (PARI) a(n)=100^n-10^n-1 \\ _Charles R Greathouse IV_, May 01 2013
%K easy,nonn
%O 0,1
%A Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003
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