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%I #56 Sep 08 2022 08:44:38
%S 0,1,4,16,65,265,1081,4410,17991,73396,299426,1221537,4983377,
%T 20330163,82938844,338356945,1380359512,5631308624,22973462017,
%U 93722435101,382349636061,1559831901918,6363483400447,25960439030624
%N Take every 5th term of Padovan sequence A000931, beginning with the second term.
%C Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n, with exactly 2 elements of each rank level above 0, for n > 0. (Uniform used in the sense of Retakh, Serconek and Wilson.) Here, we do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length. - _David Nacin_, Feb 13 2012
%D R. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.
%H Vincenzo Librandi, <a href="/A012781/b012781.txt">Table of n, a(n) for n = 0..1000</a>
%H V. Retakh, S. Serconek, and R. Wilson, <a href="http://arxiv.org/abs/1010.6295">Hilbert Series of Algebras Associated to Directed Graphs and Order Homology</a>, arXiv:1010.6295 [math.RA], 2010-2011.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4,1).
%F a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n).
%F G.f.: x*(1-x)/(1-5*x+4*x^2-x^3). - _Colin Barker_, Feb 03 2012
%t LinearRecurrence[{5, -4, 1}, {0, 1, 4}, 25] (* _Harvey P. Dale_, Jan 10 2012 *)
%o (Magma) I:=[0, 1, 4 ]; [n le 3 select I[n] else 5*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 03 2012
%o (Python)
%o def a(n, adict={0:0, 1:1, 2:4}):
%o if n in adict:
%o return adict[n]
%o adict[n]=5*a(n-1) - 4*a(n-2) + a(n-3)
%o return adict[n] # _David Nacin_, Feb 27 2012
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
%E Initial term 0 added by _Colin Barker_, Feb 03 2012
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