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a(n)=(19^n - 1)/(9*2^(3 - 2*Mod[n, 2])).
2

%I #6 Oct 21 2024 16:59:10

%S 0,1,5,381,1810,137561,653415,49659541,235882820,17927094321,

%T 85153698025,6471681049901,30740484987030,2336276859014281,

%U 11097315080317835,843395946104155461,4006130743994738440

%N a(n)=(19^n - 1)/(9*2^(3 - 2*Mod[n, 2])).

%H Ray Chandler, <a href="/A152438/b152438.txt">Table of n, a(n) for n = 0..782</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 362, 0, -361).

%t a[n_] :=(19^n - 1)/(9*2^(3 - 2*Mod[n, 2]));

%t Table[a[n], {n, 0, 30}]

%K nonn,changed

%O 0,3

%A _Roger L. Bagula_, Dec 04 2008