OFFSET
0,2
COMMENTS
Binomial transform of A079935.
Number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. 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. 96-100.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Graded Poset
Index entries for linear recurrences with constant coefficients, signature (6,-6).
FORMULA
G.f.: (1-2*x)/((1-(3-sqrt(3))*x)*(1-(3+sqrt(3))*x)) = (1-2*x)/(1-6*x+6*x^2);
a(n) = (3-sqrt(3))^n*(1/2 - 1/(2*sqrt(3))) + (3 + sqrt(3))^n*(1/2 + 1/(2*sqrt(3))).
E.g.f.: exp(3*x)*(cosh(sqrt(3)*x) + sinh(sqrt(3)*x)/sqrt(3)). - Paul Barry, Nov 20 2003
a(n) = Sum_{k=1..floor(n/2)} C(n, 2k)*3^(n-k-1). - Paul Barry, Nov 22 2003
a(n) = (((1+sqrt(3))*(3+sqrt(3))^n) - ((1-sqrt(3))*(3-sqrt(3))^n))/sqrt(12). - Al Hakanson (hawkuu(AT)gmail.com), Jun 10 2009
a(n) = Sum_{k=0..n} A117317(n,k)*2^k. - Philippe Deléham, Jan 28 2012
a(n) = 6*(a(n-1) - a(n-2)), a(0)=1, a(1)=4. - David Nacin, Feb 27 2012
G.f.: (1-2*x)/(1-6*x+6*x^2). - Colin Barker, Aug 04 2012
MATHEMATICA
LinearRecurrence[{6, -6}, {1, 4}, 60] (* David Nacin, Feb 27 2012 *)
PROG
(Python)
def a(n, adict={0:1, 1:4}):
if n in adict:
return adict[n]
adict[n]=6*a(n-1)-6*a(n-2)
return adict[n] # David Nacin, Feb 27 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jul 19 2003
STATUS
approved