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a(n) = 2^(n - 1) (n - mod(n, 2)).
0

%I #17 Feb 16 2025 08:33:51

%S 0,4,8,32,64,192,384,1024,2048,5120,10240,24576,49152,114688,229376,

%T 524288,1048576,2359296,4718592,10485760,20971520,46137344,92274688,

%U 201326592,402653184,872415232,1744830464,3758096384,7516192768,16106127360

%N a(n) = 2^(n - 1) (n - mod(n, 2)).

%C Agrees with independence number of the n-cube connected cycle graph for at least 3 <= n <= 8.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Cube-ConnectedCycleGraph.html">Cube-Connected Cycle Graph</a>.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/IndependenceNumber.html">Independence Number</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,4,-8).

%F a(n) = 2^(n - 1) (n - mod(n, 2)).

%F a(n) = 2*a(n-1) + 4*a(n-2) - 8*a(n-3).

%F G.f.: (4 x^2)/((1 - 2 x)^2 (1 + 2 x)).

%F a(n) = 2^n*A004526(n). - _R. J. Mathar_, Mar 08 2021

%F Sum_{n>=2} 1/a(n) = (3/2)*log(4/3). - _Amiram Eldar_, Apr 22 2022

%t Table[2^(n - 1) (n - Mod[n, 2]), {n, 20}]

%t LinearRecurrence[{2, 4, -8}, {0, 4, 8}, 20]

%t CoefficientList[Series[(4 x)/((1 - 2 x)^2 (1 + 2 x)), {x, 0, 20}], x]

%Y Cf. A004526.

%K nonn,easy

%O 1,2

%A _Eric W. Weisstein_, Sep 06 2017