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a(n) = (8*n + 2)^9.
1

%I #20 Apr 24 2023 02:54:39

%S 512,1000000000,198359290368,5429503678976,60716992766464,

%T 406671383849472,1953125000000000,7427658739644928,23762680013799936,

%U 66540410775079424,167619550409708032,387420489000000000,833747762130149888,1689478959002692096,3251948521156637184,5987402799531080192

%N a(n) = (8*n + 2)^9.

%H Vincenzo Librandi, <a href="/A017097/b017097.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

%F a(0)=512, a(1)=1000000000, a(2)=198359290368, a(3)=5429503678976, a(4)=60716992766464, a(5)=406671383849472, a(6)=1953125000000000, a(7)=7427658739644928, a(8)=23762680013799936, a(9)=66540410775079424, a(n)=10*a(n-1)-45*a(n-2)+120*a(n-3)-210*a(n-4)+252*a(n-5)- 210*a(n-6)+ 120*a(n-7)-45*a(n-8)+10*a(n-9)-a(n-10). - _Harvey P. Dale_, Dec 15 2011

%F From _Amiram Eldar_, Apr 24 2023: (Start)

%F a(n) = A017089(n)^9.

%F a(n) = 2^9*A016821(n).

%F Sum_{n>=0} 1/a(n) = 277*Pi^9/8455716864 + 511*zeta(9)/524288. (End)

%t (8*Range[0,30]+2)^9 (* _Harvey P. Dale_, Dec 15 2011 *)

%o (Magma) [(8*n+2)^9: n in [0..20]]; // _Vincenzo Librandi_, Jul 12 2011

%Y Cf. A013667, A016821, A017089.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_