A065330 a(n) = max { k | gcd(n, k) = k and gcd(k, 6) = 1 }.

1, 1, 1, 1, 5, 1, 7, 1, 1, 5, 11, 1, 13, 7, 5, 1, 17, 1, 19, 5, 7, 11, 23, 1, 25, 13, 1, 7, 29, 5, 31, 1, 11, 17, 35, 1, 37, 19, 13, 5, 41, 7, 43, 11, 5, 23, 47, 1, 49, 25, 17, 13, 53, 1, 55, 7, 19, 29, 59, 5, 61, 31, 7, 1, 65, 11, 67, 17, 23, 35, 71, 1, 73, 37, 25, 19, 77, 13, 79, 5, 1

Offset: 1

Author

Reinhard Zumkeller, Oct 29 2001

Keywords

• mult
• nonn

Comments

Bennett, Filaseta, & Trifonov show that if n > 8 then a(n^2 + n) > n^0.285. - Charles R Greathouse IV, May 21 2014

Examples

			a(30) = 5.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• M. A. Bennett, M. Filaseta, and O. Trifonov, On the factorization of consecutive integers, J. Reine Angew. Math. 629 (2009), pp. 171-200.

Crossrefs

Cf. A065331, A000265, A038502, A165725.

Programs

Haskell

a065330 = a038502 . a000265  -- Reinhard Zumkeller, Jul 06 2011

Magma

[n div Gcd(n, 6^n): n in [1..100]]; // Vincenzo Librandi, Feb 09 2016

Maple

A065330 := proc(n)
    local a,f,p,e ;
    a := 1 ;
    for f in ifactors(n)[2] do
        p := op(1,f) ;
        e := op(2,f) ;
        if p > 3 then
            a := a*p^e ;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, Jul 12 2012
with(padic): a := n -> n/(2^ordp(n, 2)*3^ordp(n, 3));
seq(a(n), n=1..81); # Peter Luschny, Mar 25 2014

Mathematica

f[n_] := Times @@ (First@#^Last@# & /@ Select[FactorInteger@n, First@# != 2 && First@# != 3 &]); Array[f, 81] (* Robert G. Wilson v, Aug 18 2006 *)
f[n_]:=Denominator[6^n/n];Array[f,100] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *)
Table[n / GCD[n, 6^n], {n, 100}] (* Vincenzo Librandi, Feb 09 2016 *)

PARI

a(n)=if(n<2,1,if(n%2,if(n%3,n,a(n/3)),a(n/2))) \\ Benoit Cloitre, Jun 04 2007

PARI

a(n)=n\gcd(n,6^n) \\ Not very efficient, but simple. Stanislav Sykora, Feb 08 2016

PARI

a(n)=n>>valuation(n,2)/3^valuation(n,3) \\ Charles R Greathouse IV, Mar 31 2016

Formula

a(n) * A065331(n) = n.

Multiplicative with a(2^e)=1, a(3^e)=1, a(p^e)=p^e, p>3. - Vladeta Jovovic, Nov 02 2001

A106799(n) = A001222(a(n)). - Reinhard Zumkeller, May 19 2005

a(1)=1; then a(2n)=a(n), a(2n+1)=a((2n+1)/3) if 2n+1 is divisible by 3, a(2n+1)=2n+1 otherwise. - Benoit Cloitre, Jun 04 2007

Dirichlet g.f. zeta(s-1)*(1-2^(1-s))*(1-3^(1-s))/ ( (1-2^(-s))*(1-3^(-s)) ). - R. J. Mathar, Jul 04 2011

a(n) = A038502(A000265(n)). - Reinhard Zumkeller, Jul 06 2011

a(n) = n/GCD(n,6^n). - Stanislav Sykora, Feb 08 2016

Sum_{k=1..n} a(k) ~ (1/4) * n^2. - Amiram Eldar, Oct 22 2022