A073546 -id:A073546 - cp's OEIS frontend

A227610 Number of ways 1/n can be expressed as the sum of three distinct unit fractions: 1/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

1, 6, 15, 22, 30, 45, 36, 62, 69, 84, 56, 142, 53, 124, 178, 118, 67, 191, 74, 274, 227, 145, 87, 342, 146, 162, 216, 322, 100, 461, 84, 257, 304, 199, 435, 508, 79, 204, 360, 580, 115, 587, 98, 455, 618, 192, 129, 676, 217, 417, 369, 449, 119, 573, 543, 759, 367, 240, 166, 1236, 102, 261, 857, 428, 568, 717, 115, 537, 460, 1018, 155, 1126, 112, 276, 839

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

			a(1)=1 because 1 = 1/2 + 1/3 + 1/6;
a(2)=6 because 1/2 = 1/3 + 1/7 + 1/42 = 1/3 + 1/8 + 1/24 = 1/3 + 1/9 + 1/18 = 1/3 + 1/10 + 1/15 = 1/4 + 1/5 + 1/20 = 1/4 + 1/6 + 1/12;
a(3)=15 because 1/3 = 1/x + 1/y + 1/z presented as {x,y,z}: {4,13,156}, {4,14,84}, {4,15,60}, {4,16,48}, {4,18,36}, {4,20,30}, {4,21,28}, {5,8,120}, {5,9,45}, {5,10,30}, {5,12,20}, {6,7,42}, {6,8,24}, {6,9,18}, {6,10,15}; etc.

Jud McCranie, Table of n, a(n) for n = 1..500
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions, Wolfram Library Archive.
Ron Knott, Egyptian Fractions
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A002966, A073546.
Cf. A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.
Cf. A347566, A347569.

f[n_] := Length@ Solve[1/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 70]

A272083 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses equals one.

Original entry on oeis.org

1, 6, 3, 2, 12, 6, 4, 2, 15, 10, 3, 2, 15, 12, 10, 4, 2, 15, 12, 10, 6, 4, 3, 18, 9, 3, 2, 18, 12, 9, 4, 2, 18, 12, 9, 6, 4, 3, 18, 15, 10, 9, 6, 2, 18, 15, 12, 10, 9, 4, 3, 20, 5, 4, 2, 20, 6, 5, 4, 3, 20, 12, 6, 5, 2, 20, 15, 10, 5, 4, 3, 20, 15, 12, 10, 5

Offset: 1

Peter Kagey, Apr 19 2016

			First six rows:
[1]                   because 1/1 = 1.
[6, 3, 2]             because 1/6 + 1/3 + 1/2 = 1.
[12, 6, 4, 2]         because 1/12 + 1/6 + 1/4 + 1/2 = 1.
[15, 10, 3, 2]        because 1/15 + 1/10 + 1/3 + 1/2 = 1.
[15, 12, 10, 4, 2]    because 1/15 + 1/12 + 1/10 + 1/4 + 1/2 = 1.
[15, 12, 10, 6, 4, 3] because 1/15 + 1/12 + 1/10 + 1/6 + 1/4 + 1/3 = 1.

Peter Kagey, Table of n, a(n) for n = 1..1578 (All rows with first term less than or equal to 30.)
Wikipedia, Erdős-Graham problem.

Cf. A073546, A216975, A216993, A272020, A272036, A272081, A272082.

A216975 Triangle read by rows in which row n gives the lexicographically earliest minimal sum denominators among all possible n-term Egyptian fractions with unit sum.

Original entry on oeis.org

1, 0, 0, 2, 3, 6, 2, 4, 6, 12, 3, 4, 5, 6, 20, 3, 4, 6, 10, 12, 15, 3, 4, 9, 10, 12, 15, 18, 4, 5, 6, 9, 10, 15, 18, 20, 4, 6, 8, 9, 10, 12, 15, 18, 24, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 6, 7, 8, 9, 10, 12, 14, 15, 18, 24, 28, 6, 7, 9, 10, 11, 12, 14, 15, 18, 22, 28, 33, 7, 8, 9, 10, 11, 12, 14, 15, 18, 22, 24, 28, 33

Offset: 1

Robert Price, Sep 21 2012

This sequence is the lexicographically earliest Egyptian fraction (denominators only) describing the minimal sum given in A213062.

Row 2 = [0,0] corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.

			Row 5 = [3,4,5,6,20]: lexicographically earliest minimal sum (38) denominators among 72 possible 5-term Egyptian fractions with unit sum.
1 = 1/3 + 1/4 + 1/5 + 1/6 + 1/20.
Triangle begins:
1;
0, 0;
2, 3, 6;
2, 4, 6, 12;
3, 4, 5,  6, 20;
3, 4, 6, 10, 12, 15;

Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Robert Price, Rows n = 1..24, flattened
Harry Ruderman and Paul Erdős, Problem E2427: Bounds for Egyptian fraction partitions of unity (comments), Amer. Math. Monthly, 1974 (Vol. 81), pp. 780-782.
Eric Weisstein's World of Mathematics, Egyptian Fraction
Wikipedia, Egyptian fraction
Index entries for sequences related to Egyptian fractions

Cf. A030659, A073546, A213062, A216993.

A216993 Triangle read by rows in which row n gives the lexicographically earliest denominators with the least possible maximum value among all n-term Egyptian fractions with unit sum.

Original entry on oeis.org

1, 0, 0, 2, 3, 6, 2, 4, 6, 12, 2, 4, 10, 12, 15, 3, 4, 6, 10, 12, 15, 3, 4, 9, 10, 12, 15, 18, 3, 5, 9, 10, 12, 15, 18, 20, 4, 5, 8, 9, 10, 15, 18, 20, 24, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 5, 6, 8, 9, 10, 15, 18, 20, 21, 24, 28, 4, 8, 9, 10, 12, 15, 18, 20, 21, 24, 28, 30, 4, 8, 9, 11, 12, 18, 20, 21, 22, 24, 28, 30, 33

Offset: 1

Robert Price, Sep 21 2012

This sequence is the lexicographically earliest Egyptian fraction (denominators only) describing the minimum largest denominator given in A030659.

Row 2 = [0,0] corresponds to the fact that 1 cannot be written as an Egyptian fraction with 2 (distinct) terms.

			Row 5 = [2,4,10,12,15]: lexicographically earliest denominators with the least possible maximum value (15) among 72 possible 5-term Egyptian fractions equal to 1. 1 = 1/2 + 1/4 + 1/10 + 1/12 + 1/15.
Triangle begins:
  1;
  0, 0;
  2, 3,  6;
  2, 4,  6, 12;
  2, 4, 10, 12, 15;
  3, 4,  6, 10, 12, 15;

Robert Price, Rows n = 1..24, flattened
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330.
Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Solution College Mathematics Journal, Vol. 43, No. 4, September 2012, pp. 340-342.
Harry Ruderman and Paul Erdős, Problem E2427: Bounds for Egyptian fraction partitions of unity (comments), Amer. Math. Monthly, 1974 (Vol. 81), pp. 780-782.
Eric Weisstein's World of Mathematics, Egyptian Fraction
Wikipedia, Egyptian fraction
Index entries for sequences related to Egyptian fractions

Cf. A030659, A073546, A213062, A216975, A378723.

A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

Original entry on oeis.org

0, 1, 5, 6, 9, 15, 14, 22, 21, 30, 22, 45, 17, 36, 72, 62, 22, 69, 29, 84, 77, 56, 39, 142, 48, 53, 82, 124, 30, 178, 34, 118, 94, 67, 176, 191, 29, 74, 151, 274, 37, 227, 37, 145, 220, 87, 57, 342, 80, 146, 138, 162, 39, 216, 214, 322, 134, 100, 73, 461, 31, 84, 316, 257, 197, 304, 47, 199, 166, 435, 69, 508, 34, 79, 317

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Cf. A002966, A073546.

Cf. A227610 (1/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.

f[n_] := Length@ Solve[2/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 75]

A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

Original entry on oeis.org

1, 0, 6, 0, 1, 15, 0, 1, 5, 22, 0, 0, 1, 6, 30, 0, 0, 1, 3, 9, 45, 0, 0, 1, 1, 7, 15, 36, 0, 0, 0, 2, 2, 6, 14, 62, 0, 0, 0, 1, 1, 5, 6, 22, 69, 0, 0, 0, 1, 1, 1, 5, 16, 21, 84, 0, 0, 0, 0, 1, 1, 3, 6, 15, 30, 56, 0, 0, 0, 0, 1, 4, 1, 5, 4, 15, 22, 142, 0, 0, 0, 0, 0, 1, 1, 3, 9, 9, 13, 45, 53

Offset: 1

Robert G. Wilson v, Jul 17 2013

The main diagonal is 1, 1, 1, 1, 1, 1, 1, ..., ; i.e., 1 = 1/2 + 1/3 + 1/6.

			  m\n| 1  2   3   4   5   6   7   8   9  10  11   12  13   14   15
  ---+------------------------------------------------------------
   1 | 1  6  15  22  30  45  36  62  69  84  56  142  53  124  178  A227610
   2 | 0  1   5   6   9  15  14  22  21  30  22   45  17   36   72  A227611
   3 | 0  1   1   3   7   6   6  16  15  15  13   22   8   27   30  A075785
   4 | 0  0   1   1   2   5   5   6   4   9   7   15   4   14   33  A073101
   5 | 0  0   1   2   1   1   3   5   9   6   3   12   5   18   15  A075248
   6 | 0  0   0   1   1   1   1   3   5   7   5    6   1    6    9  n/a
   7 | 0  0   0   1   1   4   1   2   2   2   2    9   6    6    7  n/a
   8 | 0  0   0   0   1   1   1   1   1   2   0    5   3    5   15  n/a
   9 | 0  0   0   0   0   1   1   3   1   1   0    3   1    2    7  n/a
  10 | 0  0   0   0   0   1   0   2   2   1   0    1   1    3    5  n/a
.
Antidiagonals are {1}, {0, 6}, {0, 1, 15}, {0, 1, 5, 22}, {0, 0, 1, 6, 30}, {0, 0, 1, 3, 9, 45}, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions, Wolfram Library Archive.
Ron Knott Egyptian Fractions
Oakland University, The Erdős Number Project
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A002966, A073546, A227610 (1/n), A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n).

f[m_, n_] := Length@ Solve[m/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Table[ f[n, m - n + 1], {m, 12}, {n, m, 1, -1}]

A347566 Number of ways 1/n can be expressed as the sum of five distinct unit fractions: 1/n = 1/p + 1/q + 1/r + 1/s + 1/t, with 0 < p < q < r < s < t.

Original entry on oeis.org

72, 2293, 15304, 47314, 112535, 190665, 368474, 577623, 925336, 1164976, 1478492, 2051830, 2240745, 3789424, 4989958, 4672559, 4467275, 7104589, 6548335, 13844524, 13580094, 10633142, 10451326, 20262957, 16621976, 18697914, 25613090, 27523671

Offset: 1

Jud McCranie, Sep 06 2021

nonn

Cf. A002966, A073546, A227610, A241883, A347569.

A347569 Number of ways 1/n can be expressed as the sum of six distinct unit fractions: 1/n = 1/p + 1/q + 1/r + 1/s + 1/t + 1/u, with 0 < p < q < r < s < t < u.

Original entry on oeis.org

2320, 244817, 3421052, 18420699, 64025680, 131223239, 431008820, 681922142

Offset: 1

Jud McCranie, Sep 06 2021

Cf. A002966, A073546, A227610, A241883, A347566.

a(7) and a(8) from Jud McCranie, Oct 15 2021

A351532 Number of integer pairs (i, j), 0 < i, j < n, such that i/(n - i) + j/(n - j) = 1.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 2, 5, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 5, 0, 0, 1, 0, 2, 1, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 7, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 2, 3

Offset: 1

Lars Blomberg, Feb 14 2022

nonn

By symmetry, if (i, j) is a solution then so is (j, i). When j=i we get n = 3i, corresponding to the solution 1/2 + 1/2 = 1. Therefore, when 3|n, a(n) > 0 and odd, otherwise a(n) >= 0 and even.
For n < 10^6, the largest term is a(720720) = 285, and 483188 terms are 0.
Other record terms: a(1081080) = 369, a(2162160) = 457, a(3243240) = 481, a(4324320) = 533, a(5405400) = 559, a(6126120) = 597. Record terms with index >= 360360 appear to occur at indices that are multiples of 180180. - Chai Wah Wu, Feb 15 2022

			For n = 3: (i, j) = (1, 1), so a(3) = 1. (1/2 + 1/2 = 1)
For n = 18: (i, j) = (3, 8), (6, 6), (8, 3), so a(18) = 3. (3/15 + 8/10 = 1/5 + 4/5 = 1)
For n = 20: (i, j) = (5, 8), (8, 5), so a(20) = 2.
For n = 36: (i, j) = (6, 16), (8, 15), (12, 12), (15, 8), (16, 6), so a(36) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A018892, A016017, A051908, A073546, A227610, A238795, A241883, A297895, A303400.

Cf. A351694, A351695 for records.

a(n)={my(x=n^2, y=2*n); sum(i=1,(n-1)/2, x-=2*n; y-=3; if(x%y==0,1,0))}

from sympy.abc import i, j
from sympy.solvers.diophantine.diophantine import diop_quadratic
def A351532(n):
    return sum(1 for d in diop_quadratic(n**2+3*i*j-2*n*(i+j)) if 0 < d[0] < n and 0 < d[1] < n) # Chai Wah Wu, Feb 15 2022

The original equation can be solved for j giving j = (n(n - 2i)) / (2n - 3i). Varying i from 1 to n-1, a(n) is given by the number of integer values of j, 0 < j < n.

Data section extended up to a(105) by Antti Karttunen, Jan 17 2025

A378723 Triangle read by rows: row n gives denominators of n distinct unit fractions (or Egyptian fractions) summing to 1, where denominators are listed in increasing order and the denominators from largest to smallest are as small as possible.

Original entry on oeis.org

1, 0, 0, 2, 3, 6, 2, 4, 6, 12, 2, 4, 10, 12, 15, 3, 4, 6, 10, 12, 15, 3, 4, 9, 10, 12, 15, 18, 4, 5, 6, 9, 10, 15, 18, 20, 4, 6, 8, 9, 10, 12, 15, 18, 24, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 6, 7, 8, 9, 10, 12, 14, 15, 18, 24, 28, 6, 7, 8, 9, 10, 14, 15, 18, 20, 24, 28, 30

Offset: 1

Sean A. Irvine, Dec 05 2024

Row 2 = [0,0] corresponds to the fact that 1 cannot be written as an Egyptian fraction with 2 (distinct) terms.
There can be more the one solution with the same smallest maximum denominator. For example, if n=8, we have:
1/3 + 1/5 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 = 1,
1/4 + 1/5 + 1/6 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 = 1.
In this sequence, the second solution is taken because 10 < 12 when reading the denominators from the right. In A216993, the first solution is taken because 3 < 4 when reading the denominators from the left.

			Triangle begins:
  1;
  0, 0;
  2, 3,  6;
  2, 4,  6, 12;
  2, 4, 10, 12, 15;
  3, 4,  6, 10, 12, 15;
  3, 4,  9, 10, 12, 15, 18;
  4, 5,  6,  9, 10, 15, 18, 20;
  4, 6,  8,  9, 10, 12, 15, 18, 24;
  5, 6,  8,  9, 10, 12, 15, 18, 20, 24;
  6, 7,  8,  9, 10, 12, 14, 15, 18, 24, 28;
  6, 7,  8,  9, 10, 14, 15, 18, 20, 24, 28, 30;
  ...

R. K. Guy, Unsolved Problems in Number Theory, 2nd Edition, page 161.

Sean A. Irvine, Table of n, a(n) for n = 1..136
K. S. Brown, Unit Fractions, smallest last term.
Sean A. Irvine Java program (github).

Cf. A073546, A216993.

cp's OEIS Frontend

A227610 Number of ways 1/n can be expressed as the sum of three distinct unit fractions: 1/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A272083 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses equals one.

Views

Author

Keywords

Examples

Links

Crossrefs

A216975 Triangle read by rows in which row n gives the lexicographically earliest minimal sum denominators among all possible n-term Egyptian fractions with unit sum.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

A216993 Triangle read by rows in which row n gives the lexicographically earliest denominators with the least possible maximum value among all n-term Egyptian fractions with unit sum.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A347566 Number of ways 1/n can be expressed as the sum of five distinct unit fractions: 1/n = 1/p + 1/q + 1/r + 1/s + 1/t, with 0 < p < q < r < s < t.

Views

Author

Keywords

Crossrefs

A347569 Number of ways 1/n can be expressed as the sum of six distinct unit fractions: 1/n = 1/p + 1/q + 1/r + 1/s + 1/t + 1/u, with 0 < p < q < r < s < t < u.

Views

Author

Keywords

Crossrefs

Extensions

A351532 Number of integer pairs (i, j), 0 < i, j < n, such that i/(n - i) + j/(n - j) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Python

Formula

Extensions

A378723 Triangle read by rows: row n gives denominators of n distinct unit fractions (or Egyptian fractions) summing to 1, where denominators are listed in increasing order and the denominators from largest to smallest are as small as possible.

Views

Author