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a(n) = 6*n^2*(2*n + 1).
1

%I #23 Sep 08 2022 08:45:57

%S 0,18,120,378,864,1650,2808,4410,6528,9234,12600,16698,21600,27378,

%T 34104,41850,50688,60690,71928,84474,98400,113778,130680,149178,

%U 169344,191250,214968,240570,268128,297714,329400

%N a(n) = 6*n^2*(2*n + 1).

%C Number of partitions of 12*n + 1 into 4 parts.

%H Vincenzo Librandi, <a href="/A190705/b190705.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 6 * A099721(n).

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=18, a(2)=120, a(3)=378. - _Harvey P. Dale_, Mar 20 2016

%e a(1)=18: there are 18 partitions of 12*1+1=13 into 4 parts:

%e [1,1,1,10], [1,1,2,9], [1,1,3,8], [1,1,4,7], [1,1,5,6],

%e [1,2,2,8], [1,2,3,7], [1,2,4,6], [1,2,5,5], [1,3,3,6],

%e [1,3,4,5], [1,4,4,4], [2,2,2,7], [2,2,3,6], [2,2,4,5],

%e [2,3,3,5], [2,3,4,4], [3,3,3,4].

%t Table[6n^2(2n + 1), {n, 0, 30}]

%t LinearRecurrence[{4,-6,4,-1},{0,18,120,378},40] (* _Harvey P. Dale_, Mar 20 2016 *)

%o (Magma) [6*n^2*(2*n+1): n in [0..40]]; // _Vincenzo Librandi_, Jun 14 2011

%o (PARI) a(n)=6*n^2*(2*n+1) \\ _Charles R Greathouse IV_, Aug 05 2013

%K nonn,easy

%O 0,2

%A _Adi Dani_, Jun 14 2011