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.
%I A021093 #62 Jun 15 2025 23:16:37
%S A021093 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 A021093 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 A021093 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 A021093 Decimal expansion of 1/89.
%C A021093 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 A021093 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 A021093 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 A021093 Generalization (since Fibonacci(j+2) = Fibonacci(j+1) + Fibonacci(j)):
%C A021093 1/89 = Sum_{j>=0} Fibonacci(j) / 10^(j+1), (this sequence)
%C A021093 1/9899 = Sum_{j>=0} Fibonacci(j) / 100^(j+1),
%C A021093 1/998999 = Sum_{j>=0} Fibonacci(j) / 1000^(j+1),
%C A021093 1/99989999 = Sum_{j>=0} Fibonacci(j) / 10000^(j+1),
%C A021093 ...
%C A021093 1 / ((10^k)^2 - (10^k)^1 - (10^k)^0) = 1 / (10^(2k) - 10^k - 1) =
%C A021093 Sum_{j>=0} Fibonacci(j) / (10^k)^(j+1), k >= 1.
%C A021093 - _Daniel Forgues_, Oct 28 2011, May 04 2013
%C A021093 Generalization (since 11^(j+1) = 11 * 11^j):
%C A021093 1/89 = Sum_{j>=0} 11^j / 100^(j+1), (this sequence)
%C A021093 1/989 = Sum_{j>=0} 11^j / 1000^(j+1),
%C A021093 1/9989 = Sum_{j>=0} 11^j / 10000^(j+1),
%C A021093 1/99989 = Sum_{j>=0} 11^j / 100000^(j+1),
%C A021093 ...
%C A021093 1 / ((10^k)^1 - 11 (10^k)^0) = 1 / (10^k - 11) =
%C A021093 Sum_{j>=0}^ 11^j / (10^k)^(j+1), k >= 2.
%C A021093 - _Daniel Forgues_, Oct 28 2011, May 04 2013
%C A021093 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 A021093 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 A021093 Jason Earls, Red Zen, Lulu Press, NY, 2007, pp. 47-48. ISBN: 978-1-4303-2017-3.
%D A021093 Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 66.
%H A021093 Mario Raso, <a href="https://hdl.handle.net/11573/1732819">Integer Sequences in Cryptography: A New Generalized Family and its Application</a>, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 10.
%H A021093 <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 A021093 Digits:=100; evalf(1/89); # _Wesley Ivan Hurt_, May 08 2014
%t A021093 RealDigits[1/89, 10, 100, -1] (* _Wesley Ivan Hurt_, May 08 2014 *)
%o A021093 (PARI) 1/89. \\ _Charles R Greathouse IV_, Dec 05 2011
%Y A021093 Cf. A000045, A001020.
%K A021093 nonn,cons,easy
%O A021093 0,4
%A A021093 _N. J. A. Sloane_, Dec 11 1996