

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).


2



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
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OFFSET

1,2


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*(n1)/2  2 for all n > 18.
See also A261878 for a similar conjecture not involving primes.


LINKS

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


EXAMPLE

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/(31)+1/(51)+1/(71)+1/(111)+1/(131) = 11/10 and 1/(prime(5)1) = 1/10 have the same fractional part.


MATHEMATICA

frac[x_]:=xFloor[x]
u[0]:=0
u[n_]:=u[n1]+1/(Prime[n]1)
S[n_]:=Table[frac[u[n]u[m1]], {m, 1, n}]
T[1]:=S[1]
T[n_]:=Union[T[n1], S[n]]
Do[Print[n, " ", Length[T[n]]], {n, 1, 60}]


CROSSREFS

Cf. A000040, A006093, A261878.
Sequence in context: A263995 A293239 A261878 * A299251 A238485 A316264
Adjacent sequences: A261990 A261991 A261992 * A261994 A261995 A261996


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Sep 09 2015


STATUS

approved



