

A021093


Decimal expansion of 1/89.


12



0, 1, 1, 2, 3, 5, 9, 5, 5, 0, 5, 6, 1, 7, 9, 7, 7, 5, 2, 8, 0, 8, 9, 8, 8, 7, 6, 4, 0, 4, 4, 9, 4, 3, 8, 2, 0, 2, 2, 4, 7, 1, 9, 1, 0, 1, 1, 2, 3, 5, 9, 5, 5, 0, 5, 6, 1, 7, 9, 7, 7, 5, 2, 8, 0, 8, 9, 8, 8, 7, 6, 4, 0, 4, 4, 9, 4, 3, 8, 2, 0, 2, 2, 4, 7, 1, 9, 1, 0, 1, 1, 2, 3, 5, 9, 5, 5, 0, 5
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OFFSET

0,4


COMMENTS

Note the strange resemblance to the Fibonacci numbers (A000045). In fact 1/89 = sum (Fibonacci(i)/10^(i+1)). (In the same way, the Lucas numbers sum up to 120/89.)  Johan Claes, Jun 11 2004
In the Red Zen reference, the decimal expansion of 1/89 and its relation to the Fibonacci sequence is discussed; also primes of the form floor(1/89 * 10^n) are given for n = 3, 5 and 631.  Jason Earls, May 28 2007
The 44digit cycle 1, 0, 1, 1, 2, 3, 5, 9, 5, 5, 0, 5, 6, 1, 7, 9, 7, 7, 5, 2, 8, 0, 8, 9, 8, 8, 7, 6, 4, 0, 4, 4, 9, 4, 3, 8, 2, 0, 2, 4, 4, 7, 1, 9 in this sequence, and the others based on eightyninths, give the successive digits of the smallest integer that is multiplied by nine when the final digit is moved from the right hand end to the left hand end.  Ian Duff, Jan 09 2009
Generalization: [since Fibonacci(i+2) = Fibonacci(i+1) + Fibonacci(i)]
1/89 = Sum_{i=0}^{infty} [Fibonacci(i) / 10^(i+1)], (this sequence)
1/9899 = Sum_{i=0}^{infty} [Fibonacci(i) / 100^(i+1)],
1/998999 = Sum_{i=0}^{infty} [Fibonacci(i) / 1000^(i+1)],
1/99989999 = Sum_{i=0}^{infty} [Fibonacci(i) / 10000^(i+1)],
...
1 / [(10^k)^2  (10^k)^1  (10^k)^0] = 1 / [10^(2k)  10^k  1] =
Sum_{i=0}^{infty} [Fibonacci(i) / (10^k)^(i+1)], k >= 1.
 Daniel Forgues, Oct 28 2011, May 04 2013
Generalization: [since 11^(i+1) = 11 * 11^(i)]
1/89 = Sum_{i=0}^{infty} [11^i / 100^(i+1)], (this sequence)
1/989 = Sum_{i=0}^{infty} [11^i / 1000^(i+1)],
1/9989 = Sum_{i=0}^{infty} [11^i / 10000^(i+1)],
1/99989 = Sum_{i=0}^{infty} [11^i / 100000^(i+1)],
...
1 / [(10^k)^1  11 (10^k)^0] = 1 / [10^k  11] =
Sum_{i=0}^{infty} [11^i / (10^k)^(i+1)], k >= 2.
 Daniel Forgues, Oct 28 2011, May 04 2013
More generally, sum_{k>=0} F(k)/x^k = x/(x^2  x  1) (= g.f. of signed Fibonacci numbers A039834, because of negative powers). This yields 10/89 for x=10. Dividing both sides by x=10 gives the constant A021093, cf first comment.  M. F. Hasler, May 07 2014
Replacing x by a power of 10 (positive or negative exponent) in an o.g.f. gives similar constants for many sequences. For example, setting x=1/1000 in (1  sqrt(1  4*x)) / (2*x) gives 1.001002005014042132... (cf. A000108).  Joerg Arndt, May 11 2014


REFERENCES

J. Earls, Red Zen, Lulu Press, NY, 2007, pp. 4748. ISBN: 9781430320173.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 66.


LINKS

Table of n, a(n) for n=0..98.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1).


MAPLE

Digits:=100; evalf(1/89); # Wesley Ivan Hurt, May 08 2014


MATHEMATICA

RealDigits[1/89, 10, 100, 1] (* Wesley Ivan Hurt, May 08 2014 *)


PROG

(PARI) 1/89. \\ Charles R Greathouse IV, Dec 05 2011


CROSSREFS

Cf. A000045, A001020.
Sequence in context: A064358 A109736 A119628 * A342444 A011026 A069805
Adjacent sequences: A021090 A021091 A021092 * A021094 A021095 A021096


KEYWORD

nonn,cons,easy


AUTHOR

N. J. A. Sloane, Dec 11 1996


STATUS

approved



