|
|
A074361
|
|
Coefficient of q^1 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=b*nu(n-1)+lambda*(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(3,1).
|
|
5
|
|
|
0, 0, 0, 3, 19, 93, 407, 1674, 6618, 25455, 95953, 356151, 1305887, 4741092, 17072484, 61055787, 217074895, 767882865, 2704365719, 9487509102, 33170122494, 115614094071, 401864286637, 1393378817259, 4820368210175
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
COMMENTS
|
Coefficient of q^0 is A006190(n+1).
|
|
LINKS
|
|
|
FORMULA
|
G.f.: (x^4+3x^3)/(1-3x-x^2)^2.
a(0)=0, a(1)=0, a(2)=0, a(3)=3, a(4)=19, a(n)=6*a(n-1)-7*a(n-2)- 6*a(n-3)- a(n-4). - Harvey P. Dale, Jan 16 2012
|
|
EXAMPLE
|
The first 6 nu polynomials are nu(0)=1, nu(1)=3, nu(2)=10, nu(3)=33+3q, nu(4)=109+19q+10q^2, nu(5)=360+93q+66q^2+36q^3+3q^4, so the coefficients of q^1 are 0,0,0,3,19,93.
|
|
MATHEMATICA
|
CoefficientList[Series[(x^4+3x^3)/(1-3x-x^2)^2, {x, 0, 30}], x] (* or *) Join[{0}, LinearRecurrence[{6, -7, -6, -1}, {0, 0, 3, 19}, 30]] (* Harvey P. Dale, Jan 16 2012 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
|
|
EXTENSIONS
|
More terms from Brent Lehman (mailbjl(AT)yahoo.com), Aug 25 2002
|
|
STATUS
|
approved
|
|
|
|