A194763 Number of k < n such that {k*2^(1/3)} > {n*2^(1/3)}, where { } = fractional part.
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, 28, 20, 12, 4, 30, 21, 12, 3, 32, 22, 12, 2, 34, 23, 12, 1, 36, 24, 12, 0, 38, 25, 12, 52, 38, 24, 10, 53, 38, 23, 8, 54, 38, 22, 6, 55, 38, 21, 4, 56, 38, 20, 2, 57, 38, 19, 76
Offset: 1
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
N:= 100: # for a(1) .. a(N) S:= [seq(frac(k*2^(1/3)),k=1..N)]: compare:= proc(x,y) local z,a,b; z:= y - x; a:= coeff(z,2^(1/3)); b:= z - a*2^(1/3); 2*a^3 + b^3 > 0 end proc: seq(nops(select(t -> compare(S[n],t),S[1..n-1])), n=1..N); # Robert Israel, Jan 31 2025
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Mathematica
r = 2^(1/3); 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}] (* A194762 *) Table[t[n], {n, 1, 100}] (* A194763 *)