A277802 The least k > 0 such that k*A004709(n) is a cube.

1, 4, 9, 2, 25, 36, 49, 3, 100, 121, 18, 169, 196, 225, 289, 12, 361, 50, 441, 484, 529, 5, 676, 98, 841, 900, 961, 1089, 1156, 1225, 6, 1369, 1444, 1521, 1681, 1764, 1849, 242, 75, 2116, 2209, 7, 20, 2601, 338, 2809, 3025, 3249, 3364, 3481, 450, 3721, 3844

Offset: 1

Author

Peter Kagey, Oct 31 2016

Keywords

• nonn
• look

Comments

This is a permutation of the cubefree numbers (A004709).

a(n) <= A004709(n)^2, with equality iff A004709(n) is squarefree. - Robert Israel, Nov 09 2016

Examples

			a(8) = 3 because 3 * A004709(8) = 3 * 9 = 3^3.
a(16) = 12 because A004709(16) = 18 = 2^1 * 3^2. The least k such that k * 2^1 * 3^2 is a cube is 2^(3 - (1 mod 3)) * 3^(3 - (2 mod 3)) = 12. - _David A. Corneth_, Nov 01 2016
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002117, A048798, A004709.

Programs

Maple

f:= proc(n) local F,E;
   F:= ifactors(n)[2];
   E:= F[..,2];
   if max(E) >= 3 then return NULL fi;
   mul(F[i,1]^(3-E[i]),i=1..nops(F));
end proc:
map(f, [$1..1000]); # Robert Israel, Nov 09 2016

Mathematica

Table[k = 1; While[! IntegerQ[(k #)^(1/3)], k++] &@ #[[n]]; k, {n, 53}] &@ Select[Range[10^4], FreeQ[FactorInteger@ #, {, k /; k > 2}] &] (* Michael De Vlieger, Nov 01 2016, after Jan Mangaldan at A004709 *)
f[p_, e_] := If[e > 2, 0, p^(Mod[-e, 3])]; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Array[s, 100], # > 0 &] (* Amiram Eldar, Feb 20 2024 *)

PARI

\\ A list of about n terms (a little more probably).
lista(n) = {n = ceil(1.21*n); my(l=List([1]), f); forprime(p=2, n, for(i=1, #l, if(l[i] * p<=n, listput(l, l[i]*p); if(l[i]*p^2<=n, listput(l,l[i]*p^2)))));listsort(l); for(i=2, #l, f=factor(l[i]); f[, 2] = vector(#f[,2],i,3-(f[i,2] % 3))~; l[i] = factorback(f));l} \\ David A. Corneth, Nov 01 2016

Python

from math import prod
from sympy import mobius, factorint, integer_nthroot
def A277802(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return prod(p**(-e%3) for p, e in factorint(m).items()) # Chai Wah Wu, Aug 05 2024

Formula

a(n) = A048798(A004709(n)).

Sum_{k=1..n} a(k) ~ c * zeta(3)^3 * n^3 / 3, where c = Product_{p prime} (1 - 1/p^2 + 1/p^5 - 1/p^6) = 0.36052971192705404983... . - Amiram Eldar, Feb 20 2024