

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].


2



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
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OFFSET

1,7


COMMENTS

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


LINKS



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 nth Harmonic Number. [Enrique Pérez Herrero, Aug 25 2009; corrected by Robert Israel, Mar 20 2016]


EXAMPLE

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).


MAPLE

N:= 100:
H:= ListTools:PartialSums([seq(1/n, n=1..N)]):
S:= ListTools:PartialSums(map(numtheory:tau, [$1..N])):


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 *)


PROG

(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


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



