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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.

%I A270599 #69 Oct 24 2018 08:20:36
%S A270599 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,
%T A270599 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,
%U A270599 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
%N 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.
%C A270599 Number of partitions of n into such odd parts that the sum of their reciprocals is one. - _Antti Karttunen_, Jul 23 2018
%C A270599 It would be nice to know whether nonzero values may occur only on n of the form 8k+1.
%H A270599 David A. Corneth, <a href="/A270599/b270599.txt">Table of n, a(n) for n = 1..370</a> (terms 1..150 from Seiichi Manyama, terms 150..273 from Antti Karttunen)
%H A270599 David A. Corneth, <a href="/A270599/a270599_1.gp.txt">Tuples up to n = 370. </a>
%H A270599 <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%F A270599 a(2*k) = 0. - _David A. Corneth_, Jul 24 2018
%e A270599 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.
%e A270599 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
%t A270599 Array[Count[IntegerPartitions[#, All, Range[1, #, 2]], _?(Total[1/#] == 1 &)] &, 70] (* _Michael De Vlieger_, Jul 26 2018 *)
%o A270599 (Ruby)
%o A270599 def f(n)
%o A270599   n - 1 + n % 2
%o A270599 end
%o A270599 def partition(n, min, max)
%o A270599   return [[]] if n == 0
%o A270599   [f(max), f(n)].min.step(min, -2).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
%o A270599 end
%o A270599 def A270599(n)
%o A270599   ary = [1]
%o A270599   (2..n).each{|m|
%o A270599     cnt = 0
%o A270599     partition(m, 2, m).each{|ary|
%o A270599       cnt += 1 if ary.inject(0){|s, i| s + 1 / i.to_r} == 1
%o A270599     }
%o A270599     ary << cnt
%o A270599   }
%o A270599   ary
%o A270599 end
%o A270599 (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
%o A270599 (PARI)
%o A270599 \\ More verbose version for computing values of a(n) for large n:
%o A270599 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));
%o A270599 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
%Y A270599 Cf. A000009, A051908.
%Y A270599 Cf. also A201644, A201646, A201647, A201648, A201649.
%K A270599 nonn
%O A270599 1,33
%A A270599 _Seiichi Manyama_, Mar 26 2016
%E A270599 Name corrected by _Antti Karttunen_, Jul 23 2018 at the suggestion of _David A. Corneth_