login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 56th year, we are closing in on 350,000 sequences, and we’ve crossed 9,700 citations (which often say “discovered thanks to the OEIS”).

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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

%I

%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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified December 8 20:37 EST 2021. Contains 349596 sequences. (Running on oeis4.)