%I #26 Sep 10 2015 06:26:02

%S 1,2,4,7,11,15,21,28,36,45,55,66,78,91,105,120,136,151,169,188,208,

%T 229,251,274,298,323,349,376,404,433,463,494,526,559,593,628,664,701,

%U 739,778,818,859,901,944,988,1033,1079,1126,1174,1223,1273,1324,1376,1429,1483,1538,1594,1651,1709,1768

%N Number of distinct fractional parts of the numbers 1/(prime(j)-1)+...+1/(prime(k)-1) with 1 <= j <= k <= n, where the fractional part of x is given by x - floor(x).

%C Conjecture: Let m be any positive integer.

%C (i) If 1/(prime(j)-1)^m+..+1/(prime(k)-1)^m and 1/(prime(s)-1)^m+...+1/(prime(t)-1)^m 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 must have m = 1 and 1/(prime(j)-1)+...+1/(prime(k)-1) = 1+1/(prime(s)-1)+...+1/(prime(t)-1); moreover, either (j,k) = (2,6) and (s,t) = (5,5), or (j,k) = (2,5) and (s,t) = (18,18), or (j,k) = (2,17) and (s,t) =(6,18).

%C (ii) If 1/(prime(j)+1)^m+..+1/(prime(k)+1)^m and 1/(prime(s)+1)^m+...+1/(prime(t)+1)^m have the same fractional part with 1 <= j <= k, 1 <= s <= t and j <= s, but the ordered pairs (j,k) and (s,t) are different, then m is equal to 1 and 1/(prime(j)+1)+...+1/(prime(k)+1) - (1/(prime(s)+1)+...+1/(prime(t)+1)) is 0 or 1; moreover, either (j,k) = (1,9) and (s,t) = (6,8), or (j,k) = (4,4) and (s,t) = (8,10), or (j,k) = (4,7) and (s,t) =(5,10), or (j,k) = (1,10) and (s,t) = (5,7).

%C (iii) For any integer d > 1, those sums 1/(prime(j)+d)^m+..+1/(prime(k)+d)^m with 1 <= j <= k have pairwise distinct fractional parts.

%C Clearly, part (i) of the conjecture implies that a(n) = n*(n-1)/2 - 2 for all n > 18.

%C See also A261878 for a similar conjecture not involving primes.

%H Zhi-Wei Sun, <a href="/A261993/b261993.txt">Table of n, a(n) for n = 1..1200</a>

%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;c35a7a46.1509">A representation problem involving unit fractions</a>, a message to Number Theory Mailing List, Sept. 9, 2015.

%e a(3) = 4 since 1/(prime(1)-1) = 1, 1/(prime(2)-1) = 1/2, 1/(prime(3)-1) = 1/4 and 1/(prime(2)-1)+1/(prime(3)-1) = 1/2+1/4 = 3/4 have pairwise distinct fractional parts.

%e a(6) = 15 since 1/(prime(1)-1) and those 1/(prime(j)-1)+...+1/(prime(k)-1) with 1 < j <= k <= 6 and (j,k) not equal to (2,6), have pairwise distinct fractional parts, but sum_{i=2..6}1/(prime(i)-1) = 1/(3-1)+1/(5-1)+1/(7-1)+1/(11-1)+1/(13-1) = 11/10 and 1/(prime(5)-1) = 1/10 have the same fractional part.

%t frac[x_]:=x-Floor[x]

%t u[0]:=0

%t u[n_]:=u[n-1]+1/(Prime[n]-1)

%t S[n_]:=Table[frac[u[n]-u[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. A000040, A006093, A261878.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Sep 09 2015