OFFSET
1,1
COMMENTS
The algorithm at A226049, with r = sqrt(2) and f(n) = 1/(n+2), gives a sum that converges to sqrt(2). The 16th partial sum differs from sqrt(2) by less than 10^(-500).
EXAMPLE
Sum of the first 12 signed Egyptian fractions: 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 - 1/67 + 1/5858 - 1/39703532 showing denominators (beginning at 10), a(1)+1, a(2)+1, a(3)+1, ...
MATHEMATICA
$MaxExtraPrecision = Infinity;
z = 9; f[n_] := 1/(n + 2); g[n_] := 1/n - 2; r = Sqrt[2]; s = 0; a[1] = NestWhile[# + 1 &, 1, ! (s += f[#]) > r &]; p = Sum[f[n], {n, 1, a[1]}]; a[2] = Floor[g[p - r]]; a[n_] := Floor[g[((-1)^n) (p - r - Sum[((-1)^k) f[a[k]], {k, 2, n - 1}])]]; Table[a[k], {k, 1, z}]
N[p - Sum[((-1)^n)*f[a[n]], {n, 2, z}] - r, 20]
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 27 2013
STATUS
approved