A270599 Number of ways to express 1 as the sum of unit fractions with odd denominators such that the sum of those denominators is n.

1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0

Offset: 1

Author

Seiichi Manyama, Mar 26 2016

Keywords

• nonn

Comments

Number of partitions of n into such odd parts that the sum of their reciprocals is one. - Antti Karttunen, Jul 23 2018

It would be nice to know whether nonzero values may occur only on n of the form 8k+1.

Examples

			1 = 1/3 + 1/3 + 1/3, the sum of denominators is 9, this is the only expression of 1 as unit fractions with odd denominators that sum to 9, so a(9)=1.
1 = 1/15 + 1/5 + 1/5 + 1/5 + 1/3 = 1/9 + 1/9 + 1/9 + 1/3 + 1/3 are the only solutions with odd denominators that sum to 33, thus a(33) = 2. - _Antti Karttunen_, Jul 24 2018
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..370 (terms 1..150 from Seiichi Manyama, terms 150..273 from Antti Karttunen)
• David A. Corneth, Tuples up to n = 370.
• Index entries for sequences related to Egyptian fractions

Crossrefs

Cf. A000009, A051908.

Cf. also A201644, A201646, A201647, A201648, A201649.

Programs

Mathematica

Array[Count[IntegerPartitions[#, All, Range[1, #, 2]], ?(Total[1/#] == 1 &)] &, 70] (* _Michael De Vlieger, Jul 26 2018 *)

PARI

A270599(n,maxfrom=n,fracsum=0) = if(!n,(1==fracsum),my(s=0, tfs, k=(maxfrom-!(maxfrom%2))); while(k >= 1, tfs = fracsum + (1/k); if(tfs > 1, return(s), s += A270599(n-k,min(k,n-k),tfs)); k -= 2); (s)); \\ Antti Karttunen, Jul 23 2018

PARI

\\ More verbose version for computing values of a(n) for large n:
A270599(n) = if(!(n%2), 0, my(s=0); forstep(k = n, 1, -2, print("A270599(", n, ") at toplevel, k=", k, " s=", s); s += A270599aux(n-k, min(k, n-k), 1/k)); (s));
A270599aux(n,maxfrom,fracsum) = if(!n,(1==fracsum),my(s=0, tfs, k=(maxfrom-!(maxfrom%2))); while(k >= 1, tfs = fracsum + (1/k); if(tfs > 1, return(s), s += A270599aux(n-k,min(k,n-k),tfs)); k -= 2); (s)); \\ Antti Karttunen, Jul 24 2018

Ruby

def f(n)
  n - 1 + n % 2
end
def partition(n, min, max)
  return [[]] if n == 0
  [f(max), f(n)].min.step(min, -2).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
end
def A270599(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

Formula

a(2*k) = 0. - David A. Corneth, Jul 24 2018

Extensions

Name corrected by Antti Karttunen, Jul 23 2018 at the suggestion of David A. Corneth