A075988 -id:A075988 - cp's OEIS frontend

A075990 Positions of records in A075988.

1, 4, 7, 10, 13, 16, 17, 19, 22, 25, 29, 31, 37, 43, 49, 57, 61, 67, 73, 81, 82, 85, 91, 97, 101, 109, 113, 115, 121, 133, 136, 141, 142, 145, 155, 157, 169, 177, 181, 193, 199, 201, 211, 226, 229, 241, 253, 261, 262, 271, 274, 277, 283, 289, 301, 307, 313, 325

Offset: 1

Clark Kimberling, Sep 28 2002

nonn

			a(5) = 13, since A075988(13) = 7 exceeds A075988(h) for h = 1, 2, ..., 12, these values being 0, 0, 0, 1, 1, 1, 3, 2, 3, 4, 4, 4.

Cf. A075988.

Name simplified by Sean A. Irvine, Mar 13 2025

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).

Original entry on oeis.org

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

Clark Kimberling, Sep 28 2002

nonn

			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.

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

Cf. A000005, A075988, A076991.

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

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)
```

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

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

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

cp's OEIS Frontend

A075990 Positions of records in A075988.

Views

Author

Keywords

Examples

Crossrefs

Extensions

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).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Formula