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A086695 a(n) = 100^n - 10^n - 1. 1
89, 9899, 998999, 99989999, 9999899999, 999998999999, 99999989999999, 9999999899999999, 999999998999999999, 99999999989999999999, 9999999999899999999999, 999999999998999999999999, 99999999999989999999999999, 9999999999999899999999999999 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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.

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.

1/a(n) = Sum_{i>=1) A000045(i-1)/10^(n*i) (see Long paper). - Michel Marcus, May 01 2013

LINKS

Table of n, a(n) for n=1..14.

C. T. Long, The Decimal Expansion of 1/89 and Related Results, The Fibonacci Quarterly, Volume 19, Number 1, February 1981

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

FORMULA

a(n) = 10^(2*n) - 10^n - 1.

PROG

(PARI) a(n)=100^n-10^n-1 \\ Charles R Greathouse IV, May 01 2013

CROSSREFS

Sequence in context: A263431 A093948 A116254 * A056568 A174758 A181681

Adjacent sequences:  A086692 A086693 A086694 * A086696 A086697 A086698

KEYWORD

easy,nonn

AUTHOR

Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003

EXTENSIONS

Offset corrected by Jon E. Schoenfield, Jun 17 2018

STATUS

approved

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Last modified October 21 08:47 EDT 2019. Contains 328292 sequences. (Running on oeis4.)