This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A157705 #10 Apr 15 2021 21:47:31
%S A157705 1,1,3,2,18,128,171,42,1,22,1348,11738,26293,17693,3271,115,13,6122,
%T A157705 228986,2070813,6324083,7397855,3361536,544247,24590,155,3,17248,
%U A157705 2413434,67035224,612026240,2274148882
%N A157705 G.f.s of the z^p coefficients of the polynomials in the GF4 denominators of A156933.
%C A157705 The formula for the PDGF4(z;n) polynomials in the GF4 denominators of A156933 can be found below.
%C A157705 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.
%C A157705 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.
%C A157705 A Maple algorithm that generates relevant GFKT4(z;p) information can be found below.
%F A157705 PDGF4(z;n) = Product_{k=0..n} (1-(2*n+1-2*k)*z)^(3*k+1) with n = 1, 2, 3, ...
%F A157705 GFKT4(z;p) = (-1)^(p)*(z^q4)*KT4(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
%F A157705 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 A157705 Some PDGF4 (z;n) are:
%e A157705 PDGF4(z; n=3) = (1-7*z)*(1-5*z)^4*(1-3*z)^7*(1-z)^10
%e A157705 PDGF4(z; n=4) = (1-9*z)*(1-7*z)^4*(1-5*z)^7*(1-3*z)^10*(1-z)^13
%e A157705 The first few GFKT4's are:
%e A157705 GFKT4(z;p=0) = 1/(1-z)
%e A157705 GFKT4(z;p=1) = -(1+3*z+2*z^2)/(1-z)^4
%e A157705 GFKT4(z;p=2) = z*(18+128*z+171*z^2+42*z^3+z^4)/(1-z)^7
%e A157705 Some KT4(z,p) polynomials are:
%e A157705 KT4(z;p=2) = 18+128*z+171*z^2+42*z^3+z^4
%e A157705 KT4(z;p=3) = 22+1348*z+11738*z^2+26293*z^3+17693*z^4+3271*z^5+115*z^6
%p A157705 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);
%Y A157705 Originator sequence A156933.
%Y A157705 See A081436 for the z^1 coefficients and A157708 for the z^2 coefficients.
%Y A157705 Row sums equal A064350 and those of A157704.
%Y A157705 Cf. A157702, A157703, A157704.
%K A157705 easy,nonn,tabf,uned
%O A157705 0,3
%A A157705 _Johannes W. Meijer_, Mar 07 2009