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a(n) = Sum_{k=0..n} 2^(n-k) * floor(k/4).
2

%I #23 Dec 22 2023 12:11:35

%S 0,0,0,0,1,3,7,15,32,66,134,270,543,1089,2181,4365,8734,17472,34948,

%T 69900,139805,279615,559235,1118475,2236956,4473918,8947842,17895690,

%U 35791387,71582781,143165569,286331145,572662298,1145324604,2290649216,4581298440

%N a(n) = Sum_{k=0..n} 2^(n-k) * floor(k/4).

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,0,1,-3,2).

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

%F a(n) = Sum_{k=0..n} floor(2^k/15).

%F a(n) = 3*a(n-1) - 2*a(n-2) + a(n-4) - 3*a(n-5) + 2*a(n-6).

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

%F a(n) = floor(2^(n+1)/15) - floor((n+1)/4).

%o (PARI) a(n, m=4, k=2) = (k^(n+1)\(k^m-1)-(n+1)\m)/(k-1);

%o (Python)

%o def A368346(n): return (1<<n+1)//15-(n+1>>2) # _Chai Wah Wu_, Dec 22 2023

%Y Partial sums of A083593.

%Y Cf. A178420, A178452, A178455, A178457, A178459, A178460, A178742.

%K nonn,easy

%O 0,6

%A _Seiichi Manyama_, Dec 22 2023