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A054365 Number of unlabeled 5-gonal cacti having n polygons.

%I A054365 #28 Jan 09 2024 09:37:12
%S A054365 1,1,1,3,17,102,811,6626,58385,532251,5011934,48344880,475982471,
%T A054365 4766639628,48434621610,498363430232,5184274255789,54451326151253,
%U A054365 576810990484823,6156943228387305,66170786572330174,715564777086617766
%N A054365 Number of unlabeled 5-gonal cacti having n polygons.
%C A054365 Also, the number of noncrossing partitions up to rotation composed of n blocks of size 5. - _Andrew Howroyd_, May 04 2018
%H A054365 Andrew Howroyd, <a href="/A054365/b054365.txt">Table of n, a(n) for n = 0..200</a>
%H A054365 Miklos Bona, Michel Bousquet, Gilbert Labelle, and Pierre Leroux, <a href="https://doi.org/10.1006/aama.1999.0665">Enumeration of m-ary cacti</a>, Advances in Applied Mathematics, 24 (2000), 22-56.
%H A054365 Luke Lippstreu, Jorge Mago, Marcus Spradlin, and Anastasia Volovich, <a href="https://arxiv.org/abs/1906.11034">Weak Separation, Positivity and Extremal Yangian Invariants</a>, arXiv:1906.11034 [hep-th], 2019.
%H A054365 <a href="/index/Ca#cacti">Index entries for sequences related to cacti</a>
%F A054365 a(n) = ((Sum_{d|n} phi(n/d)*binomial(5*d, d)) + (Sum_{d|gcd(n-1, 5)} phi(d)*binomial(5*n/d, (n-1)/d)))/(5*n) - binomial(5*n, n)/(4*n+1) for n > 0. - _Andrew Howroyd_, May 04 2018
%p A054365 with(combinat): with(numtheory): m := 5: for p from 2 to 28 do s1 := 0: s2 := 0: for d from 1 to p do if p mod d = 0 then s1 := s1+phi(p/d)*binomial(m*d, d) fi: od: for d from 1 to p-1 do if gcd(m, p-1) mod d = 0 then s2 := s2+phi(d)*binomial((p*m)/d, (p-1)/d) fi: od: printf(`%d, `, (s1+s2)/(m*p)-binomial(m*p, p)/(p*(m-1)+1)) od: # _Zerinvary Lajos_, Dec 01 2006
%t A054365 a[0] = 1;
%t A054365 a[n_] := (DivisorSum[n, EulerPhi[n/#] Binomial[5#, #]&] + DivisorSum[GCD[n - 1, 5], EulerPhi[#] Binomial[5n/#, (n-1)/#]&])/(5n) - Binomial[5n, n]/ (4n+1);
%t A054365 Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Jul 01 2018, after _Andrew Howroyd_ *)
%o A054365 (PARI) a(n) = {if(n==0, 1, (sumdiv(n, d, eulerphi(n/d)*binomial(5*d, d)) + sumdiv(gcd(n-1, 5), d, eulerphi(d)*binomial(5*n/d, (n-1)/d)))/(5*n) - binomial(5*n, n)/(4*n+1))} \\ _Andrew Howroyd_, May 04 2018
%Y A054365 Column k=5 of A303694.
%Y A054365 Cf. A054363, A054364, A303871.
%K A054365 nonn
%O A054365 0,4
%A A054365 _Simon Plouffe_
%E A054365 More terms from _Zerinvary Lajos_, Dec 01 2006