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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A121389 a(n) = 10^Fibonacci(n) - 1.

Original entry on oeis.org

0, 9, 9, 99, 999, 99999, 99999999, 9999999999999, 999999999999999999999, 9999999999999999999999999999999999, 9999999999999999999999999999999999999999999999999999999
Offset: 0

Views

Author

Rick L. Shepherd, Jul 26 2006

Keywords

Comments

Each a(n) has Fibonacci(n) (trailing) 9's. In general, if the same recurrence below is used with any a(0), a(1) >= 0, then, for all k >= 2, a(k) has the same number of trailing 9's as a(k-2) and a(k-1) have altogether (see for example A121390).

Crossrefs

Cf. A000045, A063896, A108047, A121390.

Programs

  • Mathematica
    10^Fibonacci[Range[0,10]]-1 (* Harvey P. Dale, Dec 24 2022 *)

Formula

a(n) = 10^Fibonacci(n) - 1 = 10^A000045(n) - 1 (= 9*A108047(n) for n>=1). a(0) = 0; a(1) = 9; a(n) = a(n-2)*a(n-1) + a(n-2) + a(n-1).
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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