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

89, 9899, 998999, 99989999, 9999899999, 999998999999, 99999989999999, 9999999899999999, 999999998999999999, 99999999989999999999, 9999999999899999999999, 999999999998999999999999, 99999999999989999999999999, 9999999999999899999999999999

Offset: 1

Author

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

Keywords

• easy
• nonn

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

• 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).

Crossrefs

Cf. A000045.

Programs

Mathematica

Table[100^n-10^n-1,{n,20}] (* or *) LinearRecurrence[{111,-1110,1000},{89,9899,998999},20] (* Harvey P. Dale, Nov 16 2023 *)

PARI

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

Formula

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

Extensions

Offset corrected by Jon E. Schoenfield, Jun 17 2018