A364200 Minimal number of terms of mixed-sign Egyptian fraction f such that H(n) + f is an integer, where H(n) is the n-th harmonic number.

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

Offset: 1

Author

Denis Ivanov, Jul 13 2023

Keywords

• nonn
• more

Comments

For H(n) - floor(H(n)) and ceiling(H(n)) - H(n), the shortest mixed-sign Egyptian fractions are calculated, and the smaller length of fractions is selected.

Similar to A106394 and A224820. But those sequences use the greedy algorithm, which does not guarantee the shortest length of expansion.

For 1 < n < 41, a(n) < A363937(n) only for n = 10 and n = 22.

Examples

			For n=10: H(10) = 7381/2520 = 2.928...; H(10) - floor(H(10)) = 7381/2520 - 2 = 2341/2520 = 1/2 + 1/7 + 1/8 + 1/9 + 1/20, which cannot be expressed as the sum of fewer than 5 reciprocals, and ceiling(H(10)) - H(10) = 3 - 7381/2520 = 179/2520 = 1/30 + 1/42 + 1/72, which cannot be expressed as the sum of fewer than 3 reciprocals, so A363937(10) = 3.
But 179/2520 = 1/14 - 1/2520 (a "mixed-sign Egyptian fraction"), so a(10) = 2.
		

Links

• Ron Knott, Egyptian Fractions

Crossrefs

Cf. A363937.

Programs

Mathematica

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]);
checkMixed[f_, k_, m_] := If[m == 1,
   Catch[Do[If[check[1/i - f, k], Throw@True],
     {i, Range[2, Floor[1/f]]}];
    Throw@False],
   checkMixed[f, k, m - 1]];
a[n_] := (h = HarmonicNumber[n];
  d = Min[h - Floor@h, Ceiling@h - h];
  j = 1;
  While[Not@check[d, j], j++];
  res = j;
  Do[
   If[checkMixed[d, i - m, m], res = i],
   {i, 2, j - 1}, {m, 1, i - 1}];
  res);

Formula

a(n) <= A363937(n).