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a(n) = (12*n + 11)^8.
1

%I #17 Jun 30 2026 15:46:55

%S 214358881,78310985281,2251875390625,23811286661761,146830437604321,

%T 645753531245761,2252292232139041,6634204312890625,17181861798319201,

%U 40213853471634241,86730203469006241,174859124550883201,333160561500390625,604967116961135041,1053960288888713761,1771197285652216321

%N a(n) = (12*n + 11)^8.

%H Paolo Xausa, <a href="/A017660/b017660.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).

%t (12*Range[0,20]+11)^8 (* _Harvey P. Dale_, Mar 22 2020 *)

%t (* Alternative: *)

%t LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{214358881,78310985281,2251875390625,23811286661761,146830437604321,645753531245761,2252292232139041,6634204312890625,17181861798319201},20] (* _Harvey P. Dale_, Mar 22 2020 *)

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_