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a(n) = Sum_{d|n} mu(d)*Fibonacci(2*(n/d)+1).
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%I #11 Apr 21 2026 08:55:07

%S 2,3,11,29,87,217,608,1563,4168,10854,28655,74763,196416,513616,

%T 1346169,3522981,9227463,24153416,63245984,165569166,433493816,

%U 1134874510,2971215071,7778665461,20365010985,53316094752,139583858264,365434781904,956722026039,2504729424618

%N a(n) = Sum_{d|n} mu(d)*Fibonacci(2*(n/d)+1).

%C a(n) is the number of primitive rhythms of length n (see Romero-García et al.).

%H Gonzalo Romero-García, Lama Tarsissi, Laurent Najman, and Carlos Agon, <a href="https://hal.science/hal-05582871">Counting rhythms using combinatorics on words</a>, hal-05582871, 2026. See Theorem 3 and Table 2 at pages 11-12.

%t a[n_]:=Sum[MoebiusMu[d]*Fibonacci[2(n/d)+1],{d,Divisors[n]}]; Array[a,30]

%o (PARI) a(n) = sumdiv(n, d, moebius(d)*fibonacci(2*n/d+1)); \\ _Michel Marcus_, Apr 19 2026

%Y Cf. A000045, A001519, A008683, A122367.

%K nonn,easy

%O 1,1

%A _Stefano Spezia_, Apr 18 2026