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 A117116 Denominators of an Egyptian Fraction for phi = (1+sqrt(5))/2. 26
 1, 2, 9, 145, 37986, 2345721887, 26943815937041299094, 811625643619814151937413504618770581764, 697120590223140234675813998970770820981012350673738243594006422610850113672220 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS For each term, the largest possible unit fraction is used. REFERENCES Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342. LINKS D. Eppstein, Algorithms for Egyptian Fractions Eric Weisstein's World of Mathematics, Egyptian Fraction EXAMPLE a(4) = 145 because 1/145 is the largest unit fraction less than phi - 1/1 - 1/2 - 1/9. MAPLE v:=1: for n from 1 to 10 do x:=ceil(1/((1+sqrt(5))/2-add(1/v[i], i=0..n-1))); while not x::integer do Digits:=2*Digits; x:=ceil(1/((1+sqrt(5))/2-add(1/v[i], i=0..n-1))) od; v[n]:=x; od: seq(v[i], i=0..8);  # Paolo P. Lava, May 03 2018 MATHEMATICA a = {1}; k = N[(Sqrt - 1)/2, 1000]; Do[s = Ceiling[1/k]; AppendTo[a, s]; k = k - 1/s, {n, 1, 10}]; a (* Artur Jasinski, Sep 22 2008 *) PROG (PARI) x = (1 + sqrt(5))/2 - 1; f(x, k) = if(k<1, x, f(x, k - 1) - 1/n(x, k)); n(x, k) = ceil(1/f(x, k - 1)); for(k = 0, 9, print1(if(k==0, 1, n(x, k)), ", ")) \\ Indranil Ghosh, Mar 27 2017 CROSSREFS Cf. A001622. Sequence in context: A050995 A174954 A193440 * A211935 A133468 A182948 Adjacent sequences:  A117113 A117114 A117115 * A117117 A117118 A117119 KEYWORD nonn,frac AUTHOR Paolo P. Lava & Giorgio Balzarotti, Apr 19 2006 EXTENSIONS Edited by Don Reble, Apr 21 2006 STATUS approved

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Last modified April 17 07:52 EDT 2021. Contains 343060 sequences. (Running on oeis4.)