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%I #56 Apr 26 2023 06:29:04
%S 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,
%T 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,
%U 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
%N Decimal expansion of 1/89.
%C 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
%C 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
%C The 44-digit 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 eighty-ninths, 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
%C Generalization (since Fibonacci(j+2) = Fibonacci(j+1) + Fibonacci(j)):
%C 1/89 = Sum_{j>=0} Fibonacci(j) / 10^(j+1), (this sequence)
%C 1/9899 = Sum_{j>=0} Fibonacci(j) / 100^(j+1),
%C 1/998999 = Sum_{j>=0} Fibonacci(j) / 1000^(j+1),
%C 1/99989999 = Sum_{j>=0} Fibonacci(j) / 10000^(j+1),
%C ...
%C 1 / ((10^k)^2 - (10^k)^1 - (10^k)^0) = 1 / (10^(2k) - 10^k - 1) =
%C Sum_{j>=0} Fibonacci(j) / (10^k)^(j+1), k >= 1.
%C - _Daniel Forgues_, Oct 28 2011, May 04 2013
%C Generalization (since 11^(j+1) = 11 * 11^j):
%C 1/89 = Sum_{j>=0} 11^j / 100^(j+1), (this sequence)
%C 1/989 = Sum_{j>=0} 11^j / 1000^(j+1),
%C 1/9989 = Sum_{j>=0} 11^j / 10000^(j+1),
%C 1/99989 = Sum_{j>=0} 11^j / 100000^(j+1),
%C ...
%C 1 / ((10^k)^1 - 11 (10^k)^0) = 1 / (10^k - 11) =
%C Sum_{j>=0}^ 11^j / (10^k)^(j+1), k >= 2.
%C - _Daniel Forgues_, Oct 28 2011, May 04 2013
%C 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
%C 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
%D Jason Earls, Red Zen, Lulu Press, NY, 2007, pp. 47-48. ISBN: 978-1-4303-2017-3.
%D Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 66.
%H <a href="/index/Rec#order_23">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,1).
%p Digits:=100; evalf(1/89); # _Wesley Ivan Hurt_, May 08 2014
%t RealDigits[1/89, 10, 100, -1] (* _Wesley Ivan Hurt_, May 08 2014 *)
%o (PARI) 1/89. \\ _Charles R Greathouse IV_, Dec 05 2011
%Y Cf. A000045, A001020.
%K nonn,cons,easy
%O 0,4
%A _N. J. A. Sloane_, Dec 11 1996
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