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

%I #23 Jul 02 2024 14:43:29

%S 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,

%T 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,

%U 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

%N a(n) = A026741(n)/A051712(n+1).

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

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

%C Is this a multiplicative function?

%C Multiplicative because both A026741 and A051712(n+1) are. - _Andrew Howroyd_, Jul 26 2018

%H Antti Karttunen, <a href="/A164848/b164848.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = gcd(12, n/gcd(2, n)). - _Andrew Howroyd_, Jul 26 2018

%F From _Amiram Eldar_, Oct 28 2023: (Start)

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

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

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

%p b := proc(n) n/(n+1)/(n+2) ; end: A051712 := proc(n) numer( b(n)-b(n+1)) ; end:

%p A026741 := proc(n) if type(n,'odd') then n; else n/2; fi; end:

%p A164848 := proc(n) A026741(n)/A051712(n+1) ; end: seq(A164848(n),n=1..120) ; # _R. J. Mathar_, Sep 06 2009

%t Table[GCD[12, n / GCD[2, n]], {n, 100}] (* _Vincenzo Librandi_, Jul 26 2018 *)

%o (PARI) a(n) = gcd(12, n/gcd(2, n)); \\ _Andrew Howroyd_, Jul 26 2018

%o (Magma) [Gcd(12, n div Gcd(2, n)): n in [1..100]]; // _Vincenzo Librandi_, Jul 26 2018

%Y Cf. A026741, A027642, A051712, A164555.

%K nonn,easy,mult

%O 1,3

%A _Paul Curtz_, Aug 28 2009

%E Offset set to 1 by _R. J. Mathar_, Sep 06 2009