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 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). 6
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 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 EXAMPLE 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. 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 *) PROG (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 CROSSREFS Cf. A000005, A075988, A076991. Sequence in context: A024156 A241316 A241312 * A304714 A085432 A029169 Adjacent sequences:  A075986 A075987 A075988 * A075990 A075991 A075992 KEYWORD nonn AUTHOR Clark Kimberling, Sep 28 2002 STATUS approved

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Last modified August 10 20:21 EDT 2022. Contains 356039 sequences. (Running on oeis4.)