A261993 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).

1, 2, 4, 7, 11, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 151, 169, 188, 208, 229, 251, 274, 298, 323, 349, 376, 404, 433, 463, 494, 526, 559, 593, 628, 664, 701, 739, 778, 818, 859, 901, 944, 988, 1033, 1079, 1126, 1174, 1223, 1273, 1324, 1376, 1429, 1483, 1538, 1594, 1651, 1709, 1768

Offset: 1

Author

Zhi-Wei Sun, Sep 09 2015

Keywords

• nonn

Comments

Conjecture: Let m be any positive integer.

(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).

(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).

(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.

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

See also A261878 for a similar conjecture not involving primes.

Examples

			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.
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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1200
• Zhi-Wei Sun, A representation problem involving unit fractions, a message to Number Theory Mailing List, Sept. 9, 2015.

Crossrefs

Cf. A000040, A006093, A261878.

Programs

Mathematica

frac[x_]:=x-Floor[x]
u[0]:=0
u[n_]:=u[n-1]+1/(Prime[n]-1)
S[n_]:=Table[frac[u[n]-u[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}]