|
Search: id:A069581
|
|
|
| A069581 |
|
Triangle T(m,n) giving number of unit fractions (with odd denominators) needed to represent m/n, rational (n odd), using the greedy algorithm. |
|
+0
1
|
|
| 2, 2, 3, 4, 4, 3, 4, 3, 4, 2, 1, 2, 3, 2, 3, 4, 6, 3, 2, 3, 4, 5, 4, 5, 6, 2, 3, 10, 3, 4, 3, 4, 3, 6, 9, 6, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 2, 5, 4, 5, 2, 3, 4, 7, 6, 5, 4, 5, 10, 5, 6, 6, 3, 2, 5, 4, 3, 4, 5, 4, 7, 6, 3, 4, 5, 6, 7, 6, 2, 1, 2, 3, 4, 1, 2, 3, 2, 3, 4, 5, 2, 3, 4, 3, 4, 5, 6, 6, 5, 4
(list; graph; listen)
|
|
|
OFFSET
|
3,1
|
|
|
COMMENT
|
If m/n, a rational number (n odd) is expressed as sum (1/xi), where the xi are successively chosen to be the least possible odd integers which leave a nonnegative remainder, is the sum always finite? My conjecture: odd m needs odd, even m needs even unit fractions. In the triangle: rows are the (odd) denominators, columns are 1<m<n numerators.
|
|
REFERENCES
|
R. K. Guy: Unsolved Problems in Number Theory, Second edition, Springer- Verlag, 1994, D11.
|
|
EXAMPLE
|
T(2/7) = 4 because 2/7 = 1/5 + 1/13 + 1/115 + 1/10465.
2/3; 2/5 3/5 4/5; 2/7 3/7 4/7 5/7 6/7; 2/9 3/9 4/9 5/9 6/9 7/9 8/9
2...;.2...3...4..;..4...3...4...3...4.;...2...1...2...3...2...3...4..
|
|
CROSSREFS
|
Sequence in context: A071506 A125920 A078664 this_sequence A085430 A086416 A147968
Adjacent sequences: A069578 A069579 A069580 this_sequence A069582 A069583 A069584
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Adam Kertesz (adamkertesz(AT)worldnet.att.net), Apr 24 2002
|
|
|
Search completed in 0.002 seconds
|
