A201646 -id:A201646 - cp's OEIS frontend

A239036 A set of eleven distinct positive odd numbers the sum of whose reciprocals is 1 and whose 11th term is as large as possible.

3, 5, 7, 9, 11, 13, 23, 721, 979011, 175878510309, 20622166925499467673345

Offset: 1

Arkadiusz Wesolowski, Mar 09 2014

If k is the largest number in the set of eleven distinct positive odd numbers the sum of whose reciprocals is 1, then k <= a(11).
Is there any set of eleven distinct positive odd numbers the sum of whose reciprocals is 1 and having the Egyptian number greater than 20622166925675347163457?
This is similar to the problem discussed by Curtiss (see link), but the numbers are restricted to be odd. - T. D. Noe, Mar 18 2014

			1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/23 + 1/721 + 1/979011 + 1/175878510309 + 1/20622166925499467673345 = 1.

D. R. Curtiss, On Kellogg's Diophantine problem, Amer. Math. Monthly 29 (1922), pp. 380-387.
Index entries for sequences related to Egyptian fractions

Cf. A238795, A201646.

f=0; n=3; s=11; if(s<11, break); for(t=1, s-3, print1(n, ", "); f=f+1/n; until(1>f+1/n, n=n+2)); until(numerator(1-f-1/n)==2, n=n+2); print1(n, ", "); f=f+1/n; g=2*floor((numerator(f)+1)/4)+1; until(numerator(1-f-1/g)==1, g=g+2); print1(g, ", "); f=f+1/g; print1(denominator(1-f));

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.

Original entry on oeis.org

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

Seiichi Manyama, Mar 26 2016

nonn

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.

			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

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

Cf. A000009, A051908.

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

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

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

\\ 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

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

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

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

A316113 a(n) is the least sum of a tuple containing n as an element where the denominator of the sum of reciprocals isn't divisible by any of the prime factors of n.

Original entry on oeis.org

1, 4, 9, 10, 20, 11, 35, 22, 30, 22, 66, 22, 91, 37, 28, 46, 119, 32, 152, 31, 41, 68, 184, 37, 110, 106, 93, 53, 261, 51, 279, 94, 69, 121, 59, 60, 370, 192, 97, 64, 451, 54, 430, 94, 65, 186, 517, 74, 259, 112, 122, 97, 583, 95, 92, 92, 158, 263, 767, 77, 671

Offset: 1

David A. Corneth, Jul 22 2018

nonn

			a(12) = 22 because the tuple [3, 3, 4, 12] has the sum of reciprocals 1/3 + 1/3 + 1/4 + 1/12 = 1 of which the denominator is 1 and has no common prime factors with n = 12.
a(25) = 110 because the tuple [5, 5, 25, 25, 50] has the sum of reciprocals 1/5 + 1/5 + 1/25 + 1/25 + 1/50 = 1/2 of which the denominator is 2 and has no common prime factors with n = 25.

Ray Chandler, Table of n, a(n) for n = 1..100

Cf. A051908, A201644, A201646, A201647, A201648, A201649.

More terms from Ray Chandler, Oct 02 2018

A211119 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 63, 75, 1575

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211120 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 45, 175, 525

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211121 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 27, 75, 135, 1575

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211122 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 45, 225, 315

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211123 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 63, 105, 225

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211124 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 45, 147, 1225

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

A211125 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Original entry on oeis.org

3, 5, 7, 9, 15, 21, 25, 35, 45, 135, 4725

Offset: 1

N. J. A. Sloane, Apr 02 2012

Nechemia Burshtein, All the solutions of the equation Sum_{i=1..11} 1/x_i = 1 in distinct integers of the form x_i = 3^alpha 5^beta 7^gamma, Disc. Math. (2008) Vol. 308, No. 18, 4286-4292. MR2427761 (2009e:11061)

The 17 solutions are given in A201643, A211118-A211132, A211134.

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

cp's OEIS Frontend

A239036 A set of eleven distinct positive odd numbers the sum of whose reciprocals is 1 and whose 11th term is as large as possible.

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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.

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A316113 a(n) is the least sum of a tuple containing n as an element where the denominator of the sum of reciprocals isn't divisible by any of the prime factors of n.

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A211119 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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A211120 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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A211121 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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A211122 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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A211123 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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A211124 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

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Author

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Crossrefs

A211125 One of 17 possible sets of eleven numbers of the form 3^alpha 5^beta 7^gamma whose sum of reciprocals is 1.

Views

Author

Keywords

Links

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