OFFSET
0,3
COMMENTS
See A000931 (Padovan), and the W. Lang link given there for a combinatorial interpretation and an explicit form.
a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 2's and 3's. The limiting ratio is 2/5. - Bob Selcoe, Jul 12 2013
LINKS
FORMULA
O.g.f.: 1/((1+2*x+2*x^2)*(1-2*x)) = ((3+2*x)/(1+2*x+2*x^2) + 2/(1-2*x))/5.
a(n) = (3*b(n) + 2*b(n-1) + 2^(n+1))/5, with b(n):=A108520(n), and b(-1)=0.
a(n) = 2*a(n-2) + 4*a(n-3). - Bob Selcoe, Aug 26 2014
a(n) = 2^(n+1)/5 + Re((3-i)*(-1-i)^n)/5. - Robert Israel, Aug 26 2014
5*a(n) = 2^(n+1) -A078069(n+1). - R. J. Mathar, May 14 2024
EXAMPLE
Combinatorics for (A,B)=(2,4) Padovan sequence with weighted (3,2)-Morse type code (see the W. Lang link under A000931): n=5, - -- and -- -, with weights 2^1*4^1 and 4^1*2^1, respectively, adding to 2*2*4=16=a(5).
MAPLE
seq(2^(n+1)/5 + Re((3-I)*(-1-I)^n)/5, n=0..100); # Robert Israel, Aug 26 2014
MATHEMATICA
CoefficientList[Series[1/(1 - 2*x^2 - 4*x^3), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 26 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jul 14 2010
STATUS
approved