OFFSET
1,8
COMMENTS
The Egyptian fraction representations counted by a(n) must have at least one fraction 1/n and may contain any number of larger unit fractions.
FORMULA
a(p) = 1, p prime. - Sean A. Irvine, Apr 08 2026
EXAMPLE
a(6) = 1 because {1/3,1/4,1/4,1/6} can be reorganized in a pie so that no set of consecutive fractions add to a unit fraction. (1/3,1/4,1/6,1/4) works.
a(8) = 2 because {1/3,1/4,1/6,1/8,1/8} and {1/4,1/6,1/6,1/6,1/8,1/8} can both be reorganized in a pie so that no set of consecutive fractions add to a unit fraction. For the first of these, (1/8,1/3,1/6,1/8,1/4) fails because 1/3 + 1/6 = 1/2, but (1/3,1/8,1/6,1/4,1/8) works.
a(9) does not include {1/6,1/6,1/6,1/6,1/9,1/9,1/9} because every pie made with these slices has two 1/6 pieces next to one another. These can be replaced by a 1/3.
The following table shows one solution for each possible Egyptian fraction representation with smallest element >= 1/10. (Only the denominators are shown.) In some cases, more than one solution exists (up to rotations and reflections); for example, (8,6,8,4,3) and (8,6,4,8,3) both work for the representation 1 = 1/8 + 1/8 + 1/6 + 1/4 + 1/3.
n = 1: (1)
n = 2: (2,2)
n = 3: (3,3,3)
n = 4: no solutions
n = 5: (5,5,5,5,5)
n = 6: (6,4,3,4)
n = 7: (7,7,7,7,7,7,7)
n = 8: (8,6,8,4,3)
(8,6,8,6,4,6)
n = 9: (9,9,3,9,3)
(9,9,6,4,9,4)
(9,6,9,6,9,3)
(9,9,8,9,8,6,4)
(9,8,9,8,9,8,6,8)
n = 10: (10,6,5,5,3)
(10,5,4,5,4)
(10,8,5,8,5,4)
(10,6,10,5,10,3)
(10,6,5,6,5,6)
(10,5,10,4,10,4)
(10,9,9,6,5,9,5)
(10,8,10,8,10,5,4)
(10,6,10,6,10,6,5)
(10,9,10,9,10,9,6,5)
(10,8,10,8,10,8,5,8)
CROSSREFS
KEYWORD
nonn,more,changed
AUTHOR
Gordon Hamilton, Mar 30 2026
EXTENSIONS
a(12)-a(19) from Sean A. Irvine, Apr 06 2026
a(20)-a(23) from Pontus von Brömssen, Apr 09 2026
a(1) and a(24)-a(26) from Pontus von Brömssen, Apr 11 2026
a(27) from Pontus von Brömssen, Apr 13 2026
STATUS
approved
