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 A086695 a(n) = 100^n - 10^n - 1. 0

%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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