

A021093


Decimal expansion of 1/89.


13



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_{j>=0} Fibonacci(j)/10^(j+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(j+2) = Fibonacci(j+1) + Fibonacci(j)):
1/89 = Sum_{j>=0} Fibonacci(j) / 10^(j+1), (this sequence)
1/9899 = Sum_{j>=0} Fibonacci(j) / 100^(j+1),
1/998999 = Sum_{j>=0} Fibonacci(j) / 1000^(j+1),
1/99989999 = Sum_{j>=0} Fibonacci(j) / 10000^(j+1),
...
1 / ((10^k)^2  (10^k)^1  (10^k)^0) = 1 / (10^(2k)  10^k  1) =
Sum_{j>=0} Fibonacci(j) / (10^k)^(j+1), k >= 1.
Generalization (since 11^(j+1) = 11 * 11^j):
1/89 = Sum_{j>=0} 11^j / 100^(j+1), (this sequence)
1/989 = Sum_{j>=0} 11^j / 1000^(j+1),
1/9989 = Sum_{j>=0} 11^j / 10000^(j+1),
1/99989 = Sum_{j>=0} 11^j / 100000^(j+1),
...
1 / ((10^k)^1  11 (10^k)^0) = 1 / (10^k  11) =
Sum_{j>=0}^ 11^j / (10^k)^(j+1), k >= 2.
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 with 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

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


LINKS

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).


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MATHEMATICA



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STATUS

approved



