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a(n) = (11*n + 2)^9.
1

%I #17 Dec 15 2025 20:06:24

%S 512,10604499373,2641807540224,78815638671875,922190162669056,

%T 6351461955384057,31087100296429568,119851595982618319,

%U 387420489000000000,1093685272684360901,2773078757450186752,6443858614676334363,13929745610903012864,28334269484119140625,54716887507601719296

%N a(n) = (11*n + 2)^9.

%H Paolo Xausa, <a href="/A017421/b017421.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).

%t (11Range[0,20]+2)^9 (* _Harvey P. Dale_, Jun 26 2011 *)

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_