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
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,4).
FORMULA
a(n)=a(n-1)+4a(n-5); a(n)=sum{k=0..floor(n/2), binomial(n-2k, floor(k/2))2^k}.
MATHEMATICA
CoefficientList[Series[(1+2x^2)/(1-x-4x^5), {x, 0, 40}], x] (* or *) LinearRecurrence[{1, 0, 0, 0, 4}, {1, 1, 3, 3, 3}, 50] (* Harvey P. Dale, Mar 02 2024 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 12 2004
STATUS
approved