OFFSET
0,4
COMMENTS
Row sums of the Riordan array (1,x(1+4x^2)). - Paul Barry, Jan 12 2006
6*a(n-3) is the number of distinct nonbacktracking paths of length n on a unit cube which start on a given vertex and end on the same one (if n is even) or the opposite one (if n is odd). E.g., a(7)=69 because a(7)=a(6)+4*a(4)=33+4*9=69. a(3)=5 because there are 6*a(6-3)=6*5=30 nonbacktracking paths of length 6 on a unit cube that end on the same vertex (6 is even); if we name the vertices of a unit cube ABCDEFGH in the order of x+2y+4z, such paths starting from A are ABDCGEA, ABDHFBA, ABDHFEA, ABDHGCA, ABDHGDA; the remaining 25 can be derived from these 5 reflecting them about the ABGH plane and rotating the resulting 10 around the AH axis by 120 and -120 degrees. - Michal Kaczmarczyk, Apr 24 2006
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=3, 5*a(n-3) equals the number of 5-colored compositions of n with all parts >=3, such that no adjacent parts have the same color. - Milan Janjic, Nov 27 2011
a(n+2) equals the number of words of length n on alphabet {0,1,2,3,4}, having at least two zeros between every two successive nonzero letters. - Milan Janjic, Feb 07 2015
Number of compositions of n into one sort of part 1 and four sorts of part 3 (the g.f. is 1/(1-x-4*x^3) ). - Joerg Arndt, Feb 07 2015
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 9.
Index entries for linear recurrences with constant coefficients, signature (1,0,4).
FORMULA
a(n) = Sum_{k=0..floor(n/2)} C(n-2*k, k)*4^k.
a(n) = 2^(n-1)+2^(n/2)*(cos((n+2)*arctan(sqrt(7)/7)+Pi*n/2)/4+5*sqrt(7)*sin((n+2)*arctan(sqrt(7)/7)+Pi*n/2)/28).
a(n) = Sum_{k=0..n} C(k, floor((n-k)/2))2^(n-k)*(1+(-1)^(n-k))/2. - Paul Barry, Jan 12 2006
a(n) = a(n-1) + 4*a(n-3) for n>=3, a(0)=1, a(1)=1, a(2)=1. - Michal Kaczmarczyk, Apr 24 2006
a(n) = 2^(n-1) + A110512(n)/2. - R. J. Mathar, Aug 23 2011
G.f.: Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + 4*x^2)/( x*(4*k+3 + 4*x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
a(n) = hypergeom([1/3-n/3,2/3-n/3,-n/3],[1/2-n/2,-n/2],-27). - Peter Luschny, Feb 07 2015
MAPLE
seq(add(binomial(n-2*k, k)*4^k, k=0..floor(n/3)), n=0..32); # Zerinvary Lajos, Apr 03 2007
MATHEMATICA
Table[HypergeometricPFQ[{1/3-n/3, 2/3-n/3, -n/3}, {1/2-n/2, -n/2}, -27], {n, 0, 32}] (* Peter Luschny, Feb 07 2015 *)
CoefficientList[Series[1/((1 - 2*x)*(1 + x + 2*x^2)), {x, 0, 50}], x] (* G. C. Greubel, Apr 27 2017 *)
LinearRecurrence[{1, 0, 4}, {1, 1, 1}, 40] (* Harvey P. Dale, Sep 01 2021 *)
PROG
(PARI) Vec(1/((1-2*x)*(1+x+2*x^2)) + O(x^50)) \\ Michel Marcus, Feb 07 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Nov 18 2003
STATUS
approved