login
a(n) = (4n+2)^3.
4

%I #32 Mar 08 2022 20:32:34

%S 8,216,1000,2744,5832,10648,17576,27000,39304,54872,74088,97336,

%T 125000,157464,195112,238328,287496,343000,405224,474552,551368,

%U 636056,729000,830584,941192,1061208,1191016

%N a(n) = (4n+2)^3.

%H Michael De Vlieger, <a href="/A016827/b016827.txt">Table of n, a(n) for n = 0..10000</a>

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

%F G.f.: 8*(1 + 23*x + 23*x^2 + x^3)/(1 - x)^4. - _Ilya Gutkovskiy_, Jun 21 2016

%F From _Amiram Eldar_, Jun 28 2020: (Start)

%F Sum_{n>=0} 1/a(n) = 7*zeta(3)/64.

%F Sum_{n>=0} (-1)^n/a(n) = Pi^3/256. (End)

%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Wesley Ivan Hurt_, Mar 08 2022

%t Table[(4n+2)^3,{n,0,100}] (* _Mohammad K. Azarian_, Jun 20 2016 *)

%Y Odd bisection of A016743.

%Y Subsequence of A000578.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_.