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%I #23 Mar 23 2024 10:36:25
%S 89,9899,998999,99989999,9999899999,999998999999,99999989999999,
%T 9999999899999999,999999998999999999,99999999989999999999,
%U 9999999999899999999999,999999999998999999999999,99999999999989999999999999,9999999999999899999999999999
%N a(n) = 100^n - 10^n - 1.
%C Digits of the inverses of these numbers give 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) give the x_n up to 10^n.
%C 1/a(n) = Sum_{i>=1) 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/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).
%F a(n) = 10^(2*n) - 10^n - 1.
%t Table[100^n-10^n-1,{n,20}] (* or *) LinearRecurrence[{111,-1110,1000},{89,9899,998999},20] (* _Harvey P. Dale_, Nov 16 2023 *)
%o (PARI) a(n)=100^n-10^n-1 \\ _Charles R Greathouse IV_, May 01 2013
%Y Cf. A000045.
%K easy,nonn
%O 1,1
%A Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003
%E Offset corrected by _Jon E. Schoenfield_, Jun 17 2018
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