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%I #30 Sep 11 2015 10:59:16
%S 1,2,4,7,11,15,21,28,36,45,55,64,76,89,103,118,134,151,169,187,207,
%T 228,250,273,297,322,348,375,403,432,462,493,525,558,592,627,663,700,
%U 738,777,817,858,900,943,987,1032,1078,1125,1173,1222,1272,1323,1375,1428,1482,1537,1593,1650,1708,1767
%N Number of distinct fractional parts of the sums 1/j+...+1/k with 1 <= j <= k <= n, where the fractional part of x is given by x - floor(x).
%C 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).
%C (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.
%C Clearly, part (i) of the conjecture implies that a(n) = n*(n-1)/2 - 3 for all n > 20.
%C See also A261993 for a similar conjecture involving primes.
%H Zhi-Wei Sun, <a href="/A261878/b261878.txt">Table of n, a(n) for n = 1..1200</a>
%e a(3) = 4 since the four numbers 1/1, 1/2, 1/3, 1/2+1/3 = 5/6 have pairise distinct fractional parts.
%e 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.
%t frac[x_]:=x-Floor[x]
%t H[n_]:=HarmonicNumber[n]
%t S[n_]:=Table[frac[H[n]-H[m-1]],{m,1,n}]
%t T[1]:=S[1]
%t T[n_]:=Union[T[n-1],S[n]]
%t Do[Print[n," ",Length[T[n]]],{n,1,60}]
%Y Cf. A001008, A002805, A261993.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Sep 09 2015
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