A378723 - OEIS

%I #15 Dec 07 2024 04:27:34

%S 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,

%T 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,

%U 7,8,9,10,12,14,15,18,24,28,6,7,8,9,10,14,15,18,20,24,28,30

%N 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.

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

%C There can be more the one solution with the same smallest maximum denominator. For example, if n=8, we have:

%C 1/3 + 1/5 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 = 1,

%C 1/4 + 1/5 + 1/6 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 = 1.

%C 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.

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

%H Sean A. Irvine, <a href="/A378723/b378723.txt">Table of n, a(n) for n = 1..136</a>

%H K. S. Brown, <a href="http://www.ics.uci.edu/~eppstein/numth/egypt/kterm-minden.html">Unit Fractions, smallest last term</a>.

%H Sean A. Irvine <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a378/A378723.java">Java program</a> (github).

%e Triangle begins:

%e   1;

%e   0, 0;

%e   2, 3,  6;

%e   2, 4,  6, 12;

%e   2, 4, 10, 12, 15;

%e   3, 4,  6, 10, 12, 15;

%e   3, 4,  9, 10, 12, 15, 18;

%e   4, 5,  6,  9, 10, 15, 18, 20;

%e   4, 6,  8,  9, 10, 12, 15, 18, 24;

%e   5, 6,  8,  9, 10, 12, 15, 18, 20, 24;

%e   6, 7,  8,  9, 10, 12, 14, 15, 18, 24, 28;

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

%e   ...

%Y Cf. A073546, A216993.

%K nonn,tabl,new

%O 1,4

%A _Sean A. Irvine_, Dec 05 2024