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

1,5

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.

Links

Table of n, a(n) for n=1..42.

Ron Knott, Egyptian Fractions

Formula

a(n) <= A363937(n).

Example

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.

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);

Crossrefs

Cf. A363937.

Sequence in context: A060287 A238733 A165001 * A059253 A108133 A341120

Adjacent sequences: A364197 A364198 A364199 * A364201 A364202 A364203

Keyword

nonn,more

Author

Denis Ivanov, Jul 13 2023

Status

approved