A075989 Number of k satisfying 1<=k<=n and {n/k} >= 1/2, where {n/k} is the fractional part of n/k, i.e., {n/k} = n/k - floor(n/k).

0, 0, 1, 0, 2, 1, 2, 2, 3, 2, 5, 2, 4, 5, 6, 3, 6, 6, 7, 6, 7, 6, 11, 6, 8, 9, 10, 9, 12, 9, 10, 10, 13, 12, 15, 10, 11, 14, 17, 12, 16, 13, 16, 15, 16, 17, 20, 15, 16, 18, 19, 16, 23, 20, 21, 18, 19, 20, 25, 20, 22, 23, 26, 21, 24, 21, 24, 27, 28, 25, 28, 22, 25, 28, 29, 26, 31, 30

Offset: 1

Author

Clark Kimberling, Sep 28 2002

Keywords

• nonn

Examples

			For n = 5, the fractional parts of k/n are 0, 1/2, 2/3, 1/4, 0; a(5) = 2 counts 1/2 and 2/3. A075988(5) = 1 counts 1/4 and A000005(5) = 2 counts the 0's.
		

Links

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

Crossrefs

Cf. A000005, A075988, A076991.

Programs

Maple

seq(nops(select(k -> frac(n/k) >= 1/2, [$1..n])), n=1..100); # Robert Israel, Sep 25 2016

Mathematica

Table[Count[Range@ n, k_ /; n/k - Floor[n/k] >= 1/2], {n, 78}] (* Michael De Vlieger, Sep 25 2016 *)

PARI

a(n)=n-sum(i=1,n,frac(n/i)>=1/2)

PARI

a(n)=sum(k=1,n,floor(2*n/(2*k+1))-floor(2*n/(2*k+2))) \\ Benoit Cloitre, Oct 21 2012

PARI

A075989(n)=sum(k=1,n,2*n\(2*k+1)-n\(k+1)) \\ M. F. Hasler, Oct 21 2012

Formula

a(n) + A075988(n) = n-d(n), where d(n) = A000005(n) is the number of divisors of n.

a(n) = Sum_{k=1..n} (floor(2*n/(2*k+1)) - floor(2*n/(2*k+2))); a(n) = (2*(log 2)-1)*n + O(n^(1/2)). Conjecture: a(n) = (2*(log 2)-1)*n + O(n^(1/4 + epsilon)) like for the divisor and Circle problems. - Benoit Cloitre, Oct 21 2012

Conjecture: Let f(a,b)=1, if (a+b) mod |a-b| != (a mod |a-b|)+(b mod |a-b|), and 0 otherwise. a(n) = Sum_{k=1..n-1} f(n,k). - Benedict W. J. Irwin, Sep 22 2016