OFFSET
0,3
COMMENTS
The formula for the PDGF4(z;n) polynomials in the GF4 denominators of A156933 can be found below.
The general structure of the GFKT4(z;p) that generate the z^p coefficients of the PDGF4(z;n) polynomials can also be found below. The KT4(z;p) polynomials in the numerators of the GFKT4(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT4(z;p) polynomials is 1, 3, 5, 7, 10, 13, 15, 18, 20, 23, 26, 29, 32, 34, 37, 40, 42. The differences of this sequence and that of the number of terms of the KT3(z;p), see A157704, follow a simple pattern.
A Maple algorithm that generates relevant GFKT4(z;p) information can be found below.
FORMULA
PDGF4(z;n) = Product_{k=0..n} (1-(2*n+1-2*k)*z)^(3*k+1) with n = 1, 2, 3, ...
GFKT4(z;p) = (-1)^(p)*(z^q4)*KT4(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .
EXAMPLE
Some PDGF4 (z;n) are:
PDGF4(z; n=3) = (1-7*z)*(1-5*z)^4*(1-3*z)^7*(1-z)^10
PDGF4(z; n=4) = (1-9*z)*(1-7*z)^4*(1-5*z)^7*(1-3*z)^10*(1-z)^13
The first few GFKT4's are:
GFKT4(z;p=0) = 1/(1-z)
GFKT4(z;p=1) = -(1+3*z+2*z^2)/(1-z)^4
GFKT4(z;p=2) = z*(18+128*z+171*z^2+42*z^3+z^4)/(1-z)^7
Some KT4(z,p) polynomials are:
KT4(z;p=2) = 18+128*z+171*z^2+42*z^3+z^4
KT4(z;p=3) = 22+1348*z+11738*z^2+26293*z^3+17693*z^4+3271*z^5+115*z^6
MAPLE
p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1, n1) *a(n-n1), n1=1..3*p+1): fk:=rsolve(a(n) = fn, a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*n2+1-(2*k))*z)^(3*k+1), k=0..n2): a(n2):= coeff(fz(n2), z, p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)= sort (simplify(fk)); GFKT4(p):=sum((fk)*z^k, k=0..infinity); q4:=ldegree((numer (GFKT4(p)))): KT4(p):=sort((-1)^(p)*simplify((GFKT4(p)*(1-z)^(3*p+1))/z^q4), z, ascending);
CROSSREFS
Originator sequence A156933.
KEYWORD
easy,nonn,tabf,uned
AUTHOR
Johannes W. Meijer, Mar 07 2009
STATUS
approved