login
Number of k < n such that {k*2^(1/3)} > {n*2^(1/3)}, where { } = fractional part.
3

%I #11 Jan 31 2025 11:57:17

%S 0,0,0,3,2,1,0,6,4,2,0,9,6,3,0,12,8,4,0,15,10,5,0,18,12,6,26,19,12,5,

%T 28,20,12,4,30,21,12,3,32,22,12,2,34,23,12,1,36,24,12,0,38,25,12,52,

%U 38,24,10,53,38,23,8,54,38,22,6,55,38,21,4,56,38,20,2,57,38,19,76

%N Number of k < n such that {k*2^(1/3)} > {n*2^(1/3)}, where { } = fractional part.

%H Robert Israel, <a href="/A194763/b194763.txt">Table of n, a(n) for n = 1..10000</a>

%p N:= 100: # for a(1) .. a(N)

%p S:= [seq(frac(k*2^(1/3)),k=1..N)]:

%p compare:= proc(x,y) local z,a,b;

%p z:= y - x;

%p a:= coeff(z,2^(1/3));

%p b:= z - a*2^(1/3);

%p 2*a^3 + b^3 > 0

%p end proc:

%p seq(nops(select(t -> compare(S[n],t),S[1..n-1])), n=1..N); # _Robert Israel_, Jan 31 2025

%t r = 2^(1/3); p[x_] := FractionalPart[x];

%t u[n_, k_] := If[p[k*r] <= p[n*r], 1, 0]

%t v[n_, k_] := If[p[k*r] > p[n*r], 1, 0]

%t s[n_] := Sum[u[n, k], {k, 1, n}]

%t t[n_] := Sum[v[n, k], {k, 1, n}]

%t Table[s[n], {n, 1, 100}] (* A194762 *)

%t Table[t[n], {n, 1, 100}] (* A194763 *)

%Y Cf. A194762, A194738.

%K nonn,look,changed

%O 1,4

%A _Clark Kimberling_, Sep 02 2011