%I #33 Oct 03 2023 17:18:55

%S 1,4,2,12,3,24,4,40,5,60,6,84,7,112,8,144,9,180,10,220,11,264,12,312,

%T 13,364,14,420,15,480,16,544,17,612,18,684,19,760,20,840,21,924,22,

%U 1012,23,1104,24,1200,25,1300,26,1404,27,1512,28,1624,29,1740,30

%N Denominator of the median of {1, 1/2, 1/3, ..., 1/n}.

%H Clark Kimberling, <a href="/A226725/b226725.txt">Table of n, a(n) for n = 1..2000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1).

%F a(n) = (n+1)/2 if n is odd, a(n) = n*(n/2+1) if n is even.

%F G.f.: W(0), where W(k)= 1 + 2*x*(k+2)/( 1 - x/(x + 2*(k+1)/W(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 16 2013

%F a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). - _Colin Barker_, Feb 27 2015

%F G.f.: x*(x^2-4*x-1) / ((x-1)^3*(x+1)^3). - _Colin Barker_, Feb 27 2015

%F a(n) = n^(1/2 + (-1)^n/2)*(n + 2^(1/2 + (-1)^n/2))/2. - _Wesley Ivan Hurt_, Feb 27 2015

%F a(n) = Sum_{k=0..n} (-1)^k * A061579(n,k). - _Alois P. Heinz_, Feb 10 2023

%e median{1, 1/2, 1/3, 1/4} = (1/2 + 1/3)/2 = 7/12, so that a(4) = 12.

%p A226725:=n->n^(1/2 + (-1)^n/2)*(n + 2^(1/2 + (-1)^n/2))/2: seq(A226725(n), n=1..100); # _Wesley Ivan Hurt_, Feb 27 2015

%t Denominator[Table[Median[Table[1/k, {k, n}]], {n, 120}]]

%t f[n_] := If[ OddQ@ n, Floor[(n + 1)/2], n(n/2 + 1)]; Array[f, 59] (* _Robert G. Wilson v_, Feb 27 2015 *)

%t With[{nn=30},Riffle[Range[nn],Table[2n+2n^2,{n,nn}]]] (* _Harvey P. Dale_, May 26 2019 *)

%t Riffle[Range[60],LinearRecurrence[{3,-3,1},{4,12,24},60]] (* _Harvey P. Dale_, Oct 03 2023 *)

%o (PARI) Vec(x*(x^2-4*x-1)/((x-1)^3*(x+1)^3) + O(x^100)) \\ _Colin Barker_, Feb 27 2015

%Y Cf. A093178 (numerators), A061579.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jun 19 2013

%E Formula changed for even terms by _Luca Brigada Villa_, Jun 20 2013