A117116 - OEIS

%I #35 Jul 11 2022 08:37:06

%S 1,2,9,145,37986,2345721887,26943815937041299094,

%T 811625643619814151937413504618770581764,

%U 697120590223140234675813998970770820981012350673738243594006422610850113672220

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

%C For each term, the largest possible unit fraction is used.

%H Amiram Eldar, <a href="/A117116/b117116.txt">Table of n, a(n) for n = 0..11</a>

%H Mohammad K. Azarian, <a href="http://www.jstor.org/stable/10.4169/college.math.j.42.4.329">Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958</a>, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. <a href="http://www.jstor.org/stable/10.4169/college.math.j.43.4.337">Solution</a>  published in Vol. 43, No. 4, September 2012, pp. 340-342.

%H David Eppstein, <a href="http://www.ics.uci.edu/%7Eeppstein/numth/egypt/intro.html">Algorithms for Egyptian Fractions</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>.

%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>.

%e a(4) = 145 because 1/145 is the largest unit fraction less than phi - 1/1 - 1/2 - 1/9.

%p 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

%t 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 *)

%o (PARI) x = (1 + sqrt(5))/2 - 1;

%o f(x, k) = if(k<1, x, f(x, k - 1) - 1/n(x, k));

%o n(x, k) = ceil(1/f(x, k - 1));

%o for(k = 0, 9, print1(if(k==0, 1, n(x, k)), ", ")) \\ _Indranil Ghosh_, Mar 27 2017

%Y Cf. A001622.

%K nonn,frac

%O 0,2

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Apr 19 2006

%E Edited by _Don Reble_, Apr 21 2006