A008833 Largest square dividing n.

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 16, 1, 1, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 16, 1

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• mult

Comments

The Dirichlet generating function of the arithmetic function of the largest t-th power dividing n is zeta(s)*zeta(t*s-t)/zeta(s*t), here with t=2 and in A008834 and A008835 with t=3 and t=4, respectively. - R. J. Mathar, Feb 19 2011

Links

• Daniel Forgues, Table of n, a(n) for n = 1..100000
• Henry Bottomley, Some Smarandache-type multiplicative sequences
• R. J. Mathar, Survey of Dirichlet series of multiplicative arithmetic functions arXiv:1106.4038 [math.NT], 2011-2012, Remark 16.
• Andrew Reiter, On (mod n) spirals, 2014, see also posting to Number Theory Mailing List, Mar 23 2014.
• Eric Weisstein's World of Mathematics, Square part

Crossrefs

Cf. A000188, A005563, A007913, A019554, A059897, A068310.

Cf. A008836, A358272, A001615.

Programs

Haskell

a008833 n = head $ filter ((== 0) . (mod n)) $
   reverse $ takeWhile (<= n) $ tail a000290_list
-- Reinhard Zumkeller, Nov 13 2011

Maple

A008833 := proc(n)
    expand(numtheory:-nthpow(n,2)) ;
end proc:
seq(A008833(n), n=1..100) ;

Mathematica

a[n_] := First[ Select[ Reverse[ Divisors[n]], IntegerQ[Sqrt[#]]&, 1]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Dec 12 2011 *)
f[p_, e_] := p^(2*Floor[e/2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jul 07 2020 *)

PARI

A008833(n)=n/core(n) \\ Michael B. Porter, Oct 17 2009

Python

from sympy.ntheory.factor_ import core
def A008833(n): return n//core(n) # Chai Wah Wu, Dec 30 2021

Formula

a(n) = A000188(n)^2 = n/A007913(n). Cf. A019554.

Multiplicative with a(p^e) = p^(2[e/2]). - David W. Wilson, Aug 01 2001

Dirichlet g.f.: zeta(s)*zeta(2s-2)/zeta(2s). - R. J. Mathar, Oct 31 2011

a(n) = A005563(n-1) / A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011

Sum_{k=1..n} a(k) ~ Zeta(3/2) * n^(3/2) / (3*Zeta(3)). - Vaclav Kotesovec, Feb 01 2019

a(A059897(n,k)) = A059897(a(n), a(k)). - Peter Munn, Nov 30 2019

From Ridouane Oudra, May 11 2025: (Start)

a(n) = Sum_{d|n} lambda(d)*d*psi(n/d), where lambda = A008836 and psi = A001615.

a(n) = lambda(n) * Sum_{d|n} lambda(d)*d*phi(n/d).

a(n) = A008836(n) * A358272(n). (End)