A008835 Largest 4th power dividing n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 81

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• mult

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Henry Bottomley, Some Smarandache-type multiplicative sequences.

Crossrefs

Cf. A000188, A000190, A008833, A008834, A053164, A053165.

Programs

Maple

with(numtheory): [ seq( expand(nthpow(i,4)),i=1..200) ];

Mathematica

Max@ Select[Divisors@ #, IntegerQ@ Power[#, 1/4] &] & /@ Range@ 81 (* Michael De Vlieger, Mar 18 2015 *)
f[p_, e_] := p^(e - Mod[e, 4]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 15 2023 *)

PARI

a(n) = {f = factor(n); for (i=1, #f~, f[i,2] = 4*(f[i,2]\4);); factorback(f);} \\ Michel Marcus, Mar 16 2015

Python

from math import prod
from sympy import factorint
def A008835(n): return prod(p**(e&-4) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 08 2024

Formula

a(n) = A000188(A000188(n))^4.

Multiplicative with a(p^e) = p^(4[e/4]). - Mitch Harris, Apr 19 2005

Dirichlet g.f.: zeta(s) * zeta(4s-4) / zeta(4s). - Álvar Ibeas, Feb 12 2015

Sum_{k=1..n} a(k) ~ zeta(5/4) * n^(5/4) / (5*zeta(5)) - 45*n/Pi^4. - Vaclav Kotesovec, Feb 03 2019

a(n) = n/A053165(n). - Amiram Eldar, Aug 15 2023

a(n) = A053164(n)^4. - Amiram Eldar, Sep 01 2024

Extensions

Entry improved by comments from Henry Bottomley, Feb 29 2000