A102722 Given n, sum all division remainders {n/k}, with k=1,...,n. The value a(n) is given by the floor of that sum. Note that {x}:=x-[x].

0, 0, 0, 0, 1, 0, 2, 1, 2, 2, 4, 2, 4, 4, 4, 4, 6, 4, 7, 5, 6, 7, 9, 6, 8, 9, 10, 8, 11, 8, 11, 10, 11, 13, 14, 10, 13, 14, 15, 13, 16, 13, 17, 16, 15, 17, 20, 16, 18, 17, 19, 18, 22, 20, 21, 19, 20, 22, 26, 19, 23, 25, 24, 23, 25, 23, 26, 26, 28, 26, 30, 23, 27, 29, 29, 29, 31, 29, 33

Offset: 1

Author

Carlos Alves, Feb 06 2005

Keywords

• easy
• nonn

Comments

Conjecture: a(n) ~ (1-EulerGamma)n.

Examples

			a(5) = [{5/1}+{5/2}+{5/3}+{5/4}+{5/5}]=[0+0.5+0.6666+0.25+0]=[1.4166]=1 (division by 1 or by the number itself is to be avoided).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A052488, A006218.

Cf. A000005.

Programs

Maple

N:= 100:
H:= ListTools:-PartialSums([seq(1/n,n=1..N)]):
S:= ListTools:-PartialSums(map(numtheory:-tau,[$1..N])):
seq(floor(n*H[n])-S[n],n=1..N); # Robert Israel, Mar 20 2016

Mathematica

Resto = Function[n, Sum[n/k - Floor[n/k], {k, 2, n - 1}]]; Floor[Map[Resto, Range[1, 1000]]]
Table[Floor[n*HarmonicNumber[n]] - Sum[DivisorSigma[0, k], {k, 1, n}], {n, 1, 200}] (* Enrique Pérez Herrero, Aug 25 2009 *)
Table[Floor[Sum[FractionalPart[n/k], {k, 1, n}]], {n, 1, 200}] (* Enrique Pérez Herrero, Aug 25 2009 *)

Python

from math import isqrt
from sympy import harmonic
def A102722(n): return int(n*harmonic(n))+(s:=isqrt(n))**2-(sum(n//k for k in range(1,s+1))<<1) # Chai Wah Wu, Oct 24 2023

Formula

a(n) = floor(n*H(n)) - Sum_{j=1..n} d(j), where d(n)=A000005(n) is the number of divisors of n, and H(n) is the n-th Harmonic Number. [Enrique Pérez Herrero, Aug 25 2009; corrected by Robert Israel, Mar 20 2016]

a(n) = A052488(n) - A006218(n). [Enrique Pérez Herrero, Aug 25 2009]