A216993 -id:A216993 - cp's OEIS frontend

A073546 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 largest denominator is smallest possible.

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, 6, 7, 8, 9, 10, 14, 15, 18, 20, 24, 28, 30

Offset: 3

Robert G. Wilson v, Aug 27 2002

From Sean A. Irvine, Dec 05 2024: (Start)
For better versions of this sequence see A216993 and A378723.
This sequence is retained because of the terms given in the Brown link.
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.
The definition of this sequence does not specify which of these should be retained and various rows given here are not consistent in their selection. In A378723, 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. (End)

			n=3: 2,3,6;
n=4: 2,4,6,12;
n=5: 2,4,10,12,15;
n=6: 3,4,6,10,12,15;
...

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

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.
K. S. Brown, Unit Fractions, smallest last term

Edited by Max Alekseyev, Mar 01 2018

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.

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

A073546 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 largest denominator is smallest possible.

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

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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.

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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.

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