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 (1,0,0,0,9).
FORMULA
a(n)=a(n-1)+9a(n-5); a(n)=sum{k=0..floor(n/2), binomial(n-2k, floor(k/2))3^k}.
MATHEMATICA
CoefficientList[Series[(1+3x^2)/(1-x-9x^5), {x, 0, 50}], x] (* or *) LinearRecurrence[ {1, 0, 0, 0, 9}, {1, 1, 4, 4, 4}, 50] (* Harvey P. Dale, Oct 10 2019 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 12 2004
STATUS
approved