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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.

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%I A069581 #6 Jun 21 2017 04:00:49
%S A069581 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,
%T A069581 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,
%U A069581 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
%N 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.
%C A069581 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.
%D A069581 R. K. Guy: Unsolved Problems in Number Theory, Second edition, Springer- Verlag, 1994, D11.
%e A069581 T(2/7) = 4 because 2/7 = 1/5 + 1/13 + 1/115 + 1/10465.
%e A069581 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
%e A069581 Triangle begins:
%e A069581 2;
%e A069581 2, 3, 4;
%e A069581 4, 3, 4, 3, 4;
%e A069581 2, 1, 2, 3, 2, 3, 4;
%e A069581 ...
%K A069581 nonn,tabf
%O A069581 3,1
%A A069581 _Adam Kertesz_, Apr 24 2002