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a(n) = (4n-1)*4n*(4n+1).
4

%I #29 Mar 20 2022 05:50:10

%S 0,60,504,1716,4080,7980,13800,21924,32736,46620,63960,85140,110544,

%T 140556,175560,215940,262080,314364,373176,438900,511920,592620,

%U 681384,778596,884640,999900,1124760,1259604,1404816,1560780,1727880,1906500

%N a(n) = (4n-1)*4n*(4n+1).

%H Harvey P. Dale, <a href="/A069140/b069140.txt">Table of n, a(n) for n = 0..1000</a>

%H Henry Bottomley, <a href="http://www.se16.info/js/halfarea.htm">Medians and area bisectors of a triangle</a>.

%H S. Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/NoteBooks/NoteBook2/chapterII/page2.htm">Notebook entry</a>.

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

%F Sum_{i>0} 1/a(i) = log(2)*3/4 - 1/2 = 0.019860..., which is the ratio of the area of the deltoid envelope formed by area bisectors of a triangle to the area of the triangle.

%F a(n) = 64n^3 - 4n = A007531(4n) = A069072(2n-1).

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Harvey P. Dale_, Dec 24 2012

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1/2 - log(2)/4 + log(tan(Pi/8))/(2*sqrt(2)). _Amiram Eldar_, Mar 20 2022

%e a(10) = 39*40*41 = 63960.

%t Table[64n^3-4n,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,60,504,1716},40] (* _Harvey P. Dale_, Dec 24 2012 *)

%o (PARI) a(n)=(4*n-1)*4*n*(4*n+1) \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A007531, A069072, A097321.

%K easy,nonn

%O 0,2

%A _Henry Bottomley_, Apr 08 2002