A073546 - OEIS

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

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OFFSET

3,1

COMMENTS

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)

REFERENCES

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

LINKS

Table of n, a(n) for n=3..77.

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

EXAMPLE

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;