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a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).
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%I #34 Sep 08 2022 08:44:57

%S 1,1,1,1,2,1,3,3,4,2,10,5,30,15,7,7,56,28,252,126,60,30,330,165,396,

%T 198,286,143,2002,1001,15015,15015,7280,3640,1768,884,15912,7956,3876,

%U 1938,38760,19380,406980,203490,99484,49742,1144066,572033,1961256,980628

%N a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).

%H <a href="/index/Lc#lcm">Index entries for sequences related to lcm's</a>

%F a(n) = A002944(n)/A001405(n).

%F a(n) = lcm(1..n+1)/(floor((n+3)/2)*binomial(n+1,floor((n+3)/2)). - _Paul Barry_, Jul 03 2006

%F a(n) = lcm(1,2,...,n+1) / (ceiling((n+1)/2)*binomial(n+1,floor((n+1)/2))) = A003418(n+1) / A100071(n+1). - _Max Alekseyev_, Oct 23 2015

%F a(n) = A263673(n+1) / A110654(n+1) = A180000(n+1) / A152271(n). - _Max Alekseyev_, Oct 23 2015

%F a(2*n-1) = A068553(n) = A068550(n)/n.

%e If n=10 then A002944(10)=2520, A001405(10)=252, the quotient a(10)=10.

%t Table[Apply[LCM, Binomial[n, Range[0, n]]]/Binomial[n, Floor[n/2]], {n, 0, 48}] (* _Michael De Vlieger_, Jun 29 2017 *)

%o (PARI) {A048619(n) = lcm(vector(n+1, i, i)) / binomial(n+1, (n+1)\2) / ((n+2)\2);}

%o (Magma) [Lcm([1..n+1]) div (Floor((n+3)/2)*Binomial(n+1,Floor((n+3)/2))): n in [0..50]]; // _Vincenzo Librandi_, Jul 10 2019

%Y Cf. A001405, A002944.

%K nonn,easy

%O 0,5

%A _Labos Elemer_

%E Definition corrected and a(0)=1 prepended by _Max Alekseyev_, Oct 23 2015