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

%I #19 Jul 29 2022 09:51:46

%S 1,815730721,152587890625,3512479453921,33232930569601,

%T 191707312997281,806460091894081,2724905250390625,7837433594376961,

%U 19925626416901921,45949729863572161,97906861202319841

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

%H Stefano Spezia, <a href="/A017540/b017540.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) = A001016(A017533(n)). - _Michel Marcus_, Jul 29 2022

%t (12*Range[0,20]+1)^8 (* _Harvey P. Dale_, Jul 23 2013 *)

%Y Cf. A001016, A017533.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_.