

A117116


Denominators of an Egyptian Fraction for phi = (1+sqrt(5))/2.


25



1, 2, 9, 145, 37986, 2345721887, 26943815937041299094, 811625643619814151937413504618770581764, 697120590223140234675813998970770820981012350673738243594006422610850113672220
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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. 340342.


LINKS

Table of n, a(n) for n=0..8.
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.


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
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 (djr(AT)nk.ca), Apr 21 2006


STATUS

approved



