A164848 a(n) = A026741(n)/A051712(n+1).

1, 1, 3, 2, 1, 3, 1, 4, 3, 1, 1, 6, 1, 1, 3, 4, 1, 3, 1, 2, 3, 1, 1, 12, 1, 1, 3, 2, 1, 3, 1, 4, 3, 1, 1, 6, 1, 1, 3, 4, 1, 3, 1, 2, 3, 1, 1, 12, 1, 1, 3, 2, 1, 3, 1, 4, 3, 1, 1, 6, 1, 1, 3, 4, 1, 3, 1, 2, 3, 1, 1, 12, 1, 1, 3, 2, 1, 3, 1, 4, 3, 1, 1, 6, 1, 1, 3, 4, 1, 3, 1, 2, 3, 1, 1, 12, 1, 1, 3, 2, 1, 3, 1, 4, 3

Offset

1,3

Comments

Twice connected to Bernoulli numbers A164555/A027642 via the Akiyama-Tanigawa algorithm.

Conjecture (checked for the first 3000 entries): periodic with a(n+24)=a(n).

Is this a multiplicative function?

Multiplicative because both A026741 and A051712(n+1) are. - Andrew Howroyd, Jul 26 2018

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Formula

a(n) = gcd(12, n/gcd(2, n)). - Andrew Howroyd, Jul 26 2018

From Amiram Eldar, Oct 28 2023: (Start)

Multiplicative with a(2^3) = 2^min(e-1,2), a(3^e) = 3, and a(p^e) = 1 for a prime p >= 5.

Dirichlet g.f.: zeta(s) * (1 + 1/2^(2*s) + 1/2^(3*s-1)) * (1 + 2/3^s).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 5/2. (End)

Maple

b := proc(n) n/(n+1)/(n+2) ; end: A051712 := proc(n) numer( b(n)-b(n+1)) ; end:
A026741 := proc(n) if type(n, 'odd') then n; else n/2; fi; end:
A164848 := proc(n) A026741(n)/A051712(n+1) ; end: seq(A164848(n), n=1..120) ; # R. J. Mathar, Sep 06 2009

Mathematica

Table[GCD[12, n / GCD[2, n]], {n, 100}] (* Vincenzo Librandi, Jul 26 2018 *)

Prog

(PARI) a(n) = gcd(12, n/gcd(2, n)); \\ Andrew Howroyd, Jul 26 2018
(Magma) [Gcd(12, n div Gcd(2, n)): n in [1..100]]; // Vincenzo Librandi, Jul 26 2018

Crossrefs

Cf. A026741, A027642, A051712, A164555.

Sequence in context: A205564 A036585 A260454 * A213514 A130827 A379817

Adjacent sequences: A164845 A164846 A164847 * A164849 A164850 A164851

Keyword

nonn,easy,mult

Author

Paul Curtz, Aug 28 2009

Extensions

Offset set to 1 by R. J. Mathar, Sep 06 2009

Status

approved