OFFSET
1,2
COMMENTS
Conjecture: (i) If 1/j+..+1/k and 1/s+...+1/t have the same fractional part with 0 < min{2,k} <= j <= k, 0 < min{2,t} <= s <= t and j <= s, but the ordered pairs (j,k) and (s,t) are different, then we have 1/j+...+1/k = 1+1/s+...+1/t; moreover, either (j,k) = (2,6) and (s,t) = (4,5), or (j,k) = (2,4) and (s,t) = (12,12), or (j,k) = (2,11) and (s,t) =(5,12), or (j,k) = (3,20) and (s,t) = (7,19).
(ii) Let a > b >= 0 and m > 0 be integers with gcd(a,b) = 1 < max{a,m}. Then the numbers sum_{i=j,...,k}1/(a*i-b)^m with 1 <= j <= k and (j > 1 if k > a-b = 1) have pairwise distinct fractional parts.
Clearly, part (i) of the conjecture implies that a(n) = n*(n-1)/2 - 3 for all n > 20.
See also A261993 for a similar conjecture involving primes.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..1200
EXAMPLE
a(3) = 4 since the four numbers 1/1, 1/2, 1/3, 1/2+1/3 = 5/6 have pairise distinct fractional parts.
a(6) = 15 since 1/1 and those 1/j+..+1/k with 1 < j <= k <= 6 and (j,k) not equal to (2,6), have pairwise distinct fractional parts, but 1/2+1/3+1/4+1/5+1/6 = 29/20 and 1/4+1/5 = 9/20 have the same fractional part.
MATHEMATICA
frac[x_]:=x-Floor[x]
H[n_]:=HarmonicNumber[n]
S[n_]:=Table[frac[H[n]-H[m-1]], {m, 1, n}]
T[1]:=S[1]
T[n_]:=Union[T[n-1], S[n]]
Do[Print[n, " ", Length[T[n]]], {n, 1, 60}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 09 2015
STATUS
approved