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A117116
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Denominators of an Egyptian Fraction for phi = (1+sqrt(5))/2.
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27
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1, 2, 9, 145, 37986, 2345721887, 26943815937041299094, 811625643619814151937413504618770581764, 697120590223140234675813998970770820981012350673738243594006422610850113672220
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OFFSET
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0,2
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COMMENTS
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For each term, the largest possible unit fraction is used.
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LINKS
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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.
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EXAMPLE
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a(4) = 145 because 1/145 is the largest unit fraction less than phi - 1/1 - 1/2 - 1/9.
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MAPLE
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v[0]:=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
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MATHEMATICA
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a = {1}; k = N[(Sqrt[5] - 1)/2, 1000]; Do[s = Ceiling[1/k]; AppendTo[a, s]; k = k - 1/s, {n, 1, 10}]; a (* Artur Jasinski, Sep 22 2008 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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