A051908 Number of ways to express 1 as the sum of unit fractions such that the sum of the denominators is n.

1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 1, 2, 3, 2, 2, 1, 2, 2, 2, 4, 5, 5, 2, 4, 5, 5, 9, 4, 4, 6, 4, 4, 7, 8, 4, 10, 9, 9, 11, 8, 13, 13, 15, 16, 21, 18, 16, 22, 19, 18, 30, 24, 19, 26, 28, 26, 29, 35, 29, 44, 28, 47, 48

Offset: 1

Author

Jud McCranie, Dec 16 1999

Keywords

• nonn

Comments

Also the number of partitions of n whose reciprocal sums to 1; "exact partitions". - Robert G. Wilson v, Sep 30 2009

Examples

			1 = 1/2 + 1/2, the sum of denominators is 4, and this is the only expression of 1 as unit fractions with denominator sum 4, so a(4)=1.
The a(22) = 3 partitions whose reciprocal sum is 1 are (12,4,3,3), (10,5,5,2), (8,8,4,2). - _Gus Wiseman_, Jul 16 2018
		

References

• Derrick Niederman, "Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed", a Perigee Book, Penguin Group, NY, 2009, pp. 82-83. [From Robert G. Wilson v, Sep 30 2009]

Links

• David A. Corneth, Table of n, a(n) for n = 1..200 (terms a(1)-a(86) from Jud McCranie, a(87)-a(88) from Robert G. Wilson v, a(89)-a(100) from Seiichi Manyama)
• David A. Corneth, Tuples up to n = 170
• Gus Wiseman, Sequences counting and ranking integer partitions by their reciprocal sums
• Index entries for sequences related to Egyptian fractions

Crossrefs

A028229 lists n such that a(n)=0.

Cf. A002966, A058360, A270599, A316854, A316855, A316888-A316904.

Programs

Mathematica

(* first do *) << "Combinatorica`"; (* then *) f[n_] := Block[{c = i = 0, k = PartitionsP@n, p = {n}}, While[i < k, If[1 == Plus @@ (1/p), c++ ]; i++; p = NextPartition@p]; c]; Array[f, 88] (* Robert G. Wilson v, Sep 30 2009 *)
Table[Length[Select[IntegerPartitions[n],Sum[1/m,{m,#}]==1&]],{n,30}] (* Gus Wiseman, Jul 16 2018 *)

Ruby

def partition(n, min, max)
  return [[]] if n == 0
  [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
end
def A051908(n)
  ary = [1]
  (2..n).each{|m|
    cnt = 0
    partition(m, 2, m).each{|ary|
      cnt += 1 if ary.inject(0){|s, i| s + 1 / i.to_r} == 1
    }
    ary << cnt
  }
  ary
end
p A051908(100) # Seiichi Manyama, May 31 2016

Formula

a(n) > 0 for n > 23.