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a(n) = (2*n)^8.
1

%I #34 Sep 08 2022 08:44:41

%S 0,256,65536,1679616,16777216,100000000,429981696,1475789056,

%T 4294967296,11019960576,25600000000,54875873536,110075314176,

%U 208827064576,377801998336,656100000000,1099511627776,1785793904896,2821109907456,4347792138496,6553600000000,9682651996416

%N a(n) = (2*n)^8.

%H Vincenzo Librandi, <a href="/A016748/b016748.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F a(n) = 256*A001016(n) = A001016(A005843(n)). - _Michel Marcus_, Nov 16 2013

%F G.f.: 256*x*(1+x)*(x^6 + 246*x^5 + 4047*x^4 + 11572*x^3 + 4047*x^2 + 246*x + 1) / (1-x)^9. - _R. J. Mathar_, May 01 2015

%F From _Amiram Eldar_, Oct 11 2020: (Start)

%F Sum_{n>=1} 1/a(n) = Pi^8/2419200.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 127*Pi^8/309657600. (End)

%p A016748:=n->(2*n)^8; seq(A016748(n), n=0..50); # _Wesley Ivan Hurt_, Nov 15 2013

%t Table[(2n)^8, {n,0,50}] (* _Wesley Ivan Hurt_, Nov 15 2013 *)

%t (2*Range[0,20])^8 (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36, -9, 1}, {0,256 ,65536, 1679616, 16777216, 100000000, 429981696, 1475789056, 4294967296}, 20] (* _Harvey P. Dale_, Jun 14 2016 *)

%o (Magma) [(2*n)^8: n in [0..20]]; // _Vincenzo Librandi_, Sep 05 2011

%o (PARI) vector(30, n, n--; (2*n)^8) \\ _G. C. Greubel_, Sep 15 2018

%Y Cf. A001016, A005843, A016760.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_