%I #45 Mar 07 2024 05:40:24
%S 0,10,84,286,680,1330,2300,3654,5456,7770,10660,14190,18424,23426,
%T 29260,35990,43680,52394,62196,73150,85320,98770,113564,129766,147440,
%U 166650,187460,209934,234136,260130,287980,317750,349504,383306,419220,457310,497640,540274,585276,632710,682640,735130,790244
%N a(n) = (32*n^3 - 2*n)/3.
%C This sequence alternates with the tetrahedral numbers, A000292, to create the centered octagonal pyramidal number sequence, A000447.
%H Colin Barker, <a href="/A267031/b267031.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 G.f.: 2*x*(5 + 22*x + 5*x^2)/(-1 + x)^4. - _Michael De Vlieger_, Jan 09 2016
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - _Colin Barker_, Jan 10 2016
%F From _Amiram Eldar_, Jan 04 2022: (Start)
%F Sum_{n>=1} 1/a(n) = 9*log(2)/2 - 3.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 3 - 3*(2-sqrt(2))*log(2)/4 - 3*sqrt(2)*log(sqrt(2)+2)/2. (End)
%F a(n) = binomial(4*n+1, 3). - _Michel Marcus_, Mar 05 2022
%F a(n) = 8*A000447(n) + A005843(n). - _Yasser Arath Chavez Reyes_, Mar 02 2024
%e a(4) = (32/3)*4^3 - (2/3)*4 = 680.
%t Table[(32 n^3 - 2 n)/3, {n, 0, 42}] (* or *)
%t CoefficientList[Series[(2 x (5 + 22 x + 5 x^2))/(-1 + x)^4, {x, 0, 41}], x] (* _Michael De Vlieger_, Jan 09 2016 *)
%o (Magma) [32/3*n^3-2/3*n: n in [0..35]]; // _Vincenzo Librandi_, Jan 10 2016
%o (PARI) concat(0, Vec(2*x*(5+22*x+5*x^2)/(1-x)^4 + O(x^100))) \\ _Colin Barker_, Jan 10 2016
%o (PARI) a(n) = (32*n^3 - 2*n)/3; \\ _Altug Alkan_, Jan 10 2015
%Y Cf. A000292, A000447.
%K nonn,easy
%O 0,2
%A _Peter M. Chema_, Jan 09 2016
%E More terms from _Michael De Vlieger_, Jan 09 2016
%E First term added from _Vincenzo Librandi_, Jan 10 2016