OFFSET
1,4
COMMENTS
Partial sums of A116085, which is more elementary to compute, cf. examples. Sequence A154888 has an equivalent definition except that i=j is allowed there, which yields the one-term sum 1/1 as an additional possibility, and thus A154888(n) = a(n)+1. Sequence A115855 is also about the same problem but does not require the fractions to be distinct. - M. F. Hasler, Jul 14 2016
FORMULA
A116085(n) = a(n+1) - a(n).
a(n) = Sum_{k=1..n-1} A116085(k), cf. examples. - M. F. Hasler, Jul 14 2016
EXAMPLE
a(4) = # [1/3+2/3, 1/4+3/4] = 2;
a(5) = a(4) + # [1/5+4/5, 2/5+3/5] = 2 + 2 = 4;
a(6) = a(5) + # [1/6+5/6, 1/6+1/3+1/2] = 4 + 2 = 6.
MATHEMATICA
Table[Length@ Select[Union /@ Flatten[Map[IntegerPartitions[1, {#}, Rest@ Union[Flatten@ TensorProduct[#, 1/#] &@ Range@ n /. {_Integer -> 0, k_ /; k > 1 -> 0}]] &, Range@ n], 1], Total@# == 1 &], {n, 25}] (* Michael De Vlieger, Jul 14 2016, after Robert G. Wilson v at A154888 *)
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Feb 04 2006
EXTENSIONS
a(24)-a(34) from Don Reble, Jul 13 2016
a(35)-a(41) from Giovanni Resta, Jul 15 2016
STATUS
approved