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A117116 Denominators of an Egyptian Fraction for phi = (1+sqrt(5))/2. 27
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.
LINKS
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.
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[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
MATHEMATICA
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 *)
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
KEYWORD
nonn,frac
AUTHOR
EXTENSIONS
Edited by Don Reble, Apr 21 2006
STATUS
approved

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)