

A012781


Take every 5th term of Padovan sequence A000931, beginning with the second term.


5



0, 1, 4, 16, 65, 265, 1081, 4410, 17991, 73396, 299426, 1221537, 4983377, 20330163, 82938844, 338356945, 1380359512, 5631308624, 22973462017, 93722435101, 382349636061, 1559831901918, 6363483400447, 25960439030624
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OFFSET

0,3


COMMENTS

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


REFERENCES

R. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96100.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. Retakh, S. Serconek, and R. Wilson, Hilbert Series of Algebras Associated to Directed Graphs and Order Homology, arXiv:1010.6295 [math.RA], 20102011.
Index entries for linear recurrences with constant coefficients, signature (5,4,1).


FORMULA

a(n+3) = 5*a(n+2)  4*a(n+1) + a(n).
G.f.: x*(1x)/(15*x+4*x^2x^3).  Colin Barker, Feb 03 2012


MATHEMATICA

LinearRecurrence[{5, 4, 1}, {0, 1, 4}, 25] (* Harvey P. Dale, Jan 10 2012 *)


PROG

(MAGMA) I:=[0, 1, 4 ]; [n le 3 select I[n] else 5*Self(n1)4*Self(n2)+Self(n3): n in [1..40]]; // Vincenzo Librandi, Feb 03 2012
(Python)
def a(n, adict={0:0, 1:1, 2:4}):
.if n in adict:
..return adict[n]
.adict[n]=5*a(n1)  4*a(n2) + a(n3)
.return adict[n] # David Nacin, Feb 27 2012


CROSSREFS

Sequence in context: A181879 A243872 A052927 * A132820 A165201 A026674
Adjacent sequences: A012778 A012779 A012780 * A012782 A012783 A012784


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

Initial term 0 added by Colin Barker, Feb 03 2012


STATUS

approved



