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a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-5*k-1,k).
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%I #8 Jun 12 2024 17:22:09

%S 1,1,1,1,3,5,7,12,21,34,55,92,153,251,414,686,1133,1869,3088,5103,

%T 8427,13917,22990,37975,62721,103598,171121,282646,466852,771119,

%U 1273690,2103796,3474913,5739647,9480387,15659094,25864698,42721676,70564951,116554700,192517665

%N a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-5*k-1,k).

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

%F G.f.: 1 / (1 - x/(1 - x^3)^2).

%F a(n) = a(n-1) + 2*a(n-3) - a(n-6) for n > 6.

%o (PARI) a(n) = sum(k=0, n\3, binomial(2*n-5*k-1, k));

%Y Cf. A373639.

%K nonn,easy

%O 0,5

%A _Seiichi Manyama_, Jun 12 2024