A194733 Number of k < n such that {k*r} > {n*r}, where { } = fractional part, r = (1+sqrt(5))/2 (the golden ratio).

0, 1, 0, 2, 4, 1, 4, 0, 4, 8, 2, 7, 12, 4, 10, 1, 8, 15, 4, 12, 0, 9, 18, 4, 14, 24, 8, 19, 2, 14, 26, 7, 20, 33, 12, 26, 4, 19, 34, 10, 26, 1, 18, 35, 8, 26, 44, 15, 34, 4, 24, 44, 12, 33, 0, 22, 44, 9, 32, 55, 18, 42, 4, 29, 54, 14, 40, 66, 24, 51, 8, 36, 64, 19, 48, 2, 32

Offset: 1

Author

Clark Kimberling, Sep 02 2011

Keywords

• nonn

Comments

The maximum possible value of a(n) is n-1. - Michael B. Porter, Jan 29 2012

Examples

			r = 1.618, 2r = 3.236, 3r = 4.854, and 4r = 6.472, where r=(1+sqrt(5))/2.  The fractional part of 4r is 0.472, which is less than the fractional parts of two of {r, 2r, 3r}, so a(4) = 2. - _Michael B. Porter_, Jan 29 2012
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A019587, A194738.

Programs

Haskell

a194733 n = length $ filter (nTau <) $
            map (snd . properFraction . (* tau) . fromInteger) [1..n]
   where (_, nTau) = properFraction (tau * fromInteger n)
         tau = (1 + sqrt 5) / 2
-- Reinhard Zumkeller, Jan 28 2012

Maple

Digits := 100;
A194733 := proc(n::posint)
    local a,k,phi,kfrac,nfrac ;
    phi := (1+sqrt(5))/2 ;
    a :=0 ;
    nfrac := n*phi-floor(n*phi) ;
    for k from 1 to n-1 do
        kfrac := k*phi-floor(k*phi) ;
        if evalf(kfrac-nfrac)  > 0 then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A194733(n),n=1..100) ;  # R. J. Mathar, Aug 13 2021

Mathematica

r = GoldenRatio; p[x_] := FractionalPart[x];
u[n_, k_] := If[p[k*r] <= p[n*r], 1, 0]
v[n_, k_] := If[p[k*r] > p[n*r], 1, 0]
s[n_] := Sum[u[n, k], {k, 1, n}]
t[n_] := Sum[v[n, k], {k, 1, n}]
Table[s[n], {n, 1, 100}]  (* A019587 *)
Table[t[n], {n, 1, 100}]  (* A194733 *)

Formula

a(n)+A019587(n)=n.