A363937 Minimal number of terms of an Egyptian fraction to be added to, or subtracted from, harmonic number H(n) to get an integer.

0, 1, 1, 1, 2, 2, 3, 3, 3, 3, 2, 3, 4, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 4, 5, 5, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6

Offset: 1

Author

Denis Ivanov, Jun 29 2023

Keywords

• nonn
• more

Comments

The shortest Egyptian fractions for H(n) - floor(H(n)) and ceiling(H(n)) - H(n) are calculated and the smaller length of those fractions is a(n).

Examples

			For n = 2: H(2) = 3/2 which is between 1 and 2 and they are reached by the same H(2) - 1 = 2 - H(2) = 1/2 which is 1 term, so a(2) = 1.
For n = 5: H(5) = 137/60 is between 2 and 3; going up 3 - H(5) = 1/2 + 1/6 + 1/20 is 3 terms but going down H(5) - 2 = 1/5 + 1/12 is 2 terms, so the latter is shorter and a(5) = 2 terms.
		

Links

• Ron Knott, Egyptian Fractions

Crossrefs

Cf. A281530

Programs

Mathematica

(* Thanks to Ron Knott for the algorithm. Slow for n>15. *)
check[f_, k_]:= (If[Numerator@f == 1, Return@True];
  If[k == 1, Return@False];
  Catch[Do[
    If[check[f - 1/i, k - 1], Throw@True],
    {i, Range[Ceiling[1/f],Floor[k/f]]}];
   Throw@False]
  );
a[n_]:= (h = HarmonicNumber[n];
  d = {h - Floor[h], Ceiling[h] - h};
  j = 1;
  While[Not[Or @@ (check[#, j] & /@ d)], j++]; j);

Extensions

a(31)-a(45) from Dmitry Petukhov, Jul 24 2023