login
G.f.s of the z^p coefficients of the polynomials in the GF1 denominators of A156921.
7

%I #10 Apr 15 2021 21:47:12

%S 1,1,1,7,26,7,3,166,951,951,166,3,263,8999,59637,108602,59637,8999,

%T 263,174,33124,848555,6062651,15477896,15477896,6062651,848555,33124,

%U 174,45,66963,5856626,122966782,920090513

%N G.f.s of the z^p coefficients of the polynomials in the GF1 denominators of A156921.

%C The formula for the PDGF1(z;n) polynomials in the GF1 denominators of A156921 can be found below.

%C The general structure of the GFKT1(z;p) that generate the z^p coefficients of the PDGF1(z; n) polynomials can also be found below. The KT1(z;p) polynomials in the numerators of the GFKT1(z;p) have a nice symmetrical structure.

%C The sequence of the number of terms of the first few KT1(z;p) polynomials is: 1, 2, 3, 6, 7, 10, 13, 14, 17, 20, 23, 24, 27, 30, 33, 36, 37, 40. The first differences follow a simple pattern. The positions of the 1's follow the Lazy Caterer's sequence A000124.

%C A Maple algorithm that generates relevant GFKT1(z;p) information can be found below.

%F PDGF1(z;n) = Product_{m=1..n} (1-(2*m-1)*z)^(n+1-m) with n = 1, 2, 3, ... .

%F GFKT1(z;p) = (-1)^(p)*(z^q1)*KT1(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ... .

%F 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, ... .

%e Some PDGF1 (z;n) are:

%e PDGF1(z;n=3) = (1-5*z)*(1-3*z)^2*(1-z)^3

%e PDGF1(z;n=4) = ((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)

%e The first few GFKT1's are:

%e GFKT1(z;p=0) = 1/(1-z)

%e GFKT1(z;p=1) = -z*(1+z)/(1-z)^4

%e GFKT1(z;p=2) = z^2*(7+26*z+7*z^2)/(1-z)^7

%e Some KT1(z;p) polynomials are:

%e KT1(z;p=2) = 7+26*z+7*z^2

%e KT1(z;p=3) = 3+166*z+951*z^2+951*z^3+166*z^4+3*z^5

%e KT1(z;p=4) = 263+8999*z+59637*z^2+108602*z^3+59637*z^4+8999*z^5+263*z^6

%p 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*m-1)*z)^(n2+1-m),m=1..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)); GFKT1(p):=(sum(fk*z^k,k=0..infinity)); q1:=ldegree((numer(GFKT1(p)))): KT1(p):=sort((-1)^p*simplify((GFKT1(p))*(1-z)^(3*p+1)/z^q1),z, ascending);

%Y Originator sequence A156921.

%Y See A000330 for the z^1 coefficients and A157706 for the z^2 coefficients.

%Y Row sums equal A052502.

%Y Cf. A157703, A157704, A157705.

%K easy,nonn,tabf,uned

%O 0,4

%A _Johannes W. Meijer_, Mar 07 2009