

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



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;
refs;
listen;
history;
text;
internal format)



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 A326846
Adjacent sequences: A069578 A069579 A069580 * A069582 A069583 A069584


KEYWORD

nonn,tabf


AUTHOR

Adam Kertesz, Apr 24 2002


STATUS

approved



