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).

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

Author

Zhi-Wei Sun, Sep 11 2015

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A058312, A058313, A261878.

Programs

Mathematica

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}]