OFFSET
1,6
COMMENTS
a(n) <= A111233(n).
a(n) <= floor(Sum_{k=1..n} 1/k) = A055980(n). - Joerg Arndt, Oct 13 2012
a(n) <= 4 for n <= 94, a(n) <= 5 for n <= 257, a(n) <= 6 for n <= 689. That is because if there is a term 1/a with p dividing a for a prime p, then there must be another term 1/b with p dividing b. Hence, not all terms from 1/1 to 1/n can be summed up. Cf. the "filter" function in my Sage script. - Manfred Scheucher, Aug 17 2015
REFERENCES
P. Erdos and R. L. Graham, Old and new problems and results in combinatorial number theory, Université de Genève, 1980.
LINKS
Manfred Scheucher, Sage Script
H. Yokota, On number of integers representable as sums of unit fractions, Canad. Math. Bull. Vol. 33 (2), 1990.
H. Yokota, On Number of Integers Representable as a Sum of Unit Fractions, II, Journal of Number Theory 67, 162-169, 1997.
EXAMPLE
1, 1/2 + 1/3 + 1/6 = 1 and 1 + 1/2 + 1/3 + 1/6 = 2 are integers, but only 2 of them are distinct, so a(6)=2.
a(24)=3 because 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 + 1/24 = 3 and Sum_{k=1..n} 1/k < 4 for all n <= 30.
a(65)=4 because the sum of the reciprocals of the integers in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 27, 28, 30, 33, 35, 36, 40, 42, 45, 48, 52, 54, 56, 60, 63, 65 } is 4 and Sum_{k=1..n} 1/k < 5 for all n <= 82. - Jon E. Schoenfield, Apr 30 2018
PROG
(PARI) ufr(n) = {tab = []; for (i=1, 2^n - 1, vb = binary(i); while(length(vb) < n, vb = concat(0, vb); );; val = sum(j=1, length(vb), vb[j]/j); if (denominator(val) == 1, tab = concat(tab, val); ); ); return (length(Set(tab))); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Oct 11 2012
EXTENSIONS
a(25)-a(46) from Manfred Scheucher, Aug 17 2015
a(47)-a(87) from Jon E. Schoenfield, Apr 30 2018
STATUS
approved