OFFSET
0,3
COMMENTS
The expansion of (1+kx^2)/(1-x-k^2*x^5) satisfies the recurrence a(n)=a(n-1)+k^2*a(n-5),a(0)=1,a(1)=1,a(2)=k+1,a(3)=k+1,a(4)=k+1, with a(n)=sum{k=0..floor(n/2), binomial(n-2k,floor(k/2))r^k}.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,16).
FORMULA
a(n)=a(n-1)+16a(n-5); a(n)=sum{k=0..floor(n/2), binomial(n-2k, floor(k/2))4^k}.
a(0)=1, a(1)=1, a(2)=5, a(3)=5, a(4)=5, a(n)=a(n-1)+16*a(n-5). - Harvey P. Dale, Jun 15 2014
MATHEMATICA
CoefficientList[Series[(1+4x^2)/(1-x-16x^5), {x, 0, 40}], x] (* or *) LinearRecurrence[{1, 0, 0, 0, 16}, {1, 1, 5, 5, 5}, 40] (* Harvey P. Dale, Jun 15 2014 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 12 2004
STATUS
approved