A261878 -id:A261878 - cp's OEIS frontend

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

Zhi-Wei Sun, Sep 09 2015

nonn

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.

			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.

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.

Cf. A000040, A006093, A261878.

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}]

A262136 Number of distinct fractional parts of the numbers Sum_{i=j..k} (-1)^i/i with 1 <= j <= k <= n, where the fractional part of x is given by x - floor(x).

Original entry on oeis.org

1, 2, 4, 7, 11, 14, 20, 27, 35, 44, 54, 64, 76, 89, 103, 118, 134, 151, 169, 186, 206, 227, 249, 272, 296, 321, 347, 374, 402, 430, 460, 491, 523, 556, 590, 625, 661, 698, 736, 775, 815, 854, 896, 939, 983, 1028, 1074, 1121, 1169, 1218, 1268, 1319, 1371, 1424, 1478, 1532, 1588, 1645, 1703, 1762

Offset: 1

Zhi-Wei Sun, Sep 11 2015

nonn

Note that (-1)^n/n+(-1)^(n+1)/(n+1) = (-1)^n/(n*(n+1)) for any n > 0.
Conjecture: (i) Suppose that Sum_{i=j..k} (-1)^i/i and Sum_{r=s..t} (-1)^r/r with 0 < min{2,k} <= j <= k, 0 < min{2,t} <= s <= t and j <= s have the same fractional part, but the ordered pairs (j,k) and (s,t) are different. Then Sum_{i=j..k} (-1)^i/i = Sum_{r=s..t} (-1)^r/r. Moreover, if j is odd, then j > 1, k = j*(j+1) and (s,t) = (j+2,j*(j+1)-1); if j is even, then either (k = j+1 and s = t = j*(j+1)), or (k = j*(j+1)-1 and (s,t) = (j+2,j*(j+1))).
(ii) Let a > b >= 0 and m > 0 be integers with gcd(a,b) = 1 < max{a,m}. For each r = 0,1, the numbers Sum_{i=j..k} (-1)^(i-r*j)/(a*i-b)^m with 1 <= j <= k and (j > 1 if k > a-b = 1) have pairwise distinct fractional parts.
This is an analog of the conjecture in A261878. Part (i) of the conjecture implies that a(n) = n*(n-1)/2 + 2 - floor((sqrt(4n+1)-1)/2) - floor((sqrt(4n+1)-1)/4) for all n > 1.

			a(6) = 14 since the sums (-1)^j/j+...+(-1)^k/k with 0 < min{k,2} <= j <= k <= 6 and (j,k) different from (4,6) and (6,6) have pairwise distinct fractional parts, but (-1)^6/6 = (-1)^2/2+(-1)^3/3 and 1/4-1/5+1/6 = 1/2-1/3+1/4-1/5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1000

Cf. A058312, A058313, A261878.

frac[x_]:=x-Floor[x]
u[0]:=0
u[n_]:=u[n-1]+(-1)^n/n
S[n_]:=Table[frac[u[n]-u[m-1]],{m,Min[2,n],n}]
T[1]:=S[1]
T[n_]:=Union[T[n-1],S[n]]
Do[Print[n," ",Length[T[n]]],{n,1,60}]

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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).

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