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.

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

Offset

3,1

Comments

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.

Links

Table of n, a(n) for n=3..105.

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
Triangle begins:
2;
2, 3, 4;
4, 3, 4, 3, 4;
2, 1, 2, 3, 2, 3, 4;
...

Crossrefs

Sequence in context: A328388 A325954 A243503 * A274061 A284009 A339695

Adjacent sequences: A069578 A069579 A069580 * A069582 A069583 A069584

Keyword

nonn,tabf

Author

Adam Kertesz, Apr 24 2002

Status

approved