|
%I #18 Apr 01 2020 04:58:04
%S 2,5,141,68575,32089377154,1644444237306316731482,
%T 65593236350142721999718859354569312622907814
%N Egyptian fraction for square root of 1/2.
%H Jinyuan Wang, <a href="/A069139/b069139.txt">Table of n, a(n) for n = 0..10</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>
%F a(n) = ceiling(1/(1/sqrt(2) - Sum_{i=0..n-1} 1/a(i))).
%e a(3) = 68575 since sqrt(1/2) = 0.707106781186..., 1/2 + 1/5 + 1/141 + 1/68574 = 0.707106781368... (which is too much) and 1/2 + 1/5 + 1/141 + 1/68575 = 0.707106781155... (which is not too much).
%t a = {}; k = N[1/Sqrt[2], 1000]; Do[s = Ceiling[1/k]; AppendTo[a, s]; k = k - 1/s, {n, 1, 10}]; a (* _Artur Jasinski_, Sep 22 2008 *)
%Y Cf. A006487, A010503 (sqrt(1/2)).
%K nonn
%O 0,1
%A _Henry Bottomley_, Apr 08 2002
| 