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 A157704 #10 Apr 15 2021 21:47:24
%S A157704 1,1,5,32,186,132,10,56,2814,17834,27324,11364,1078,10,48,17988,
%T A157704 494720,3324209,7526484,6382271,2004296,203799,4580,5,16,72210,
%U A157704 7108338,146595355,1025458635,2957655028,3828236468
%N A157704 G.f.s of the z^p coefficients of the polynomials in the GF3 denominators of A156927.
%C A157704 The formula for the PDGF3(z;n) polynomials in the GF3 denominators of A156927 can be found below.
%C A157704 The general structure of the GFKT3(z;p) that generate the z^p coefficients of the PDGF3(z; n) polynomials can also be found below. The KT3(z;p) polynomials in the numerators of the GFKT3(z; p) have a nice symmetrical structure.
%C A157704 The sequence of the number of terms of the first few KT3(z;p) polynomials is 1, 2, 4, 7, 10, 13, 14, 17, 20, 23, 26, 29, 32, 34, 36, 39, 42. The differences of this sequence and that of the number of terms of the KT4(z;p), see A157705, follow a simple pattern.
%C A157704 A Maple algorithm that generates relevant GFKT3(z;p) information can be found below.
%F A157704 PDGF3(z;n) = Product_{k=0..n} (1-(k+1)*z)^(1+3*k) with n = 1, 2, 3, ...
%F A157704 GFKT3(z;p) = (-1)^(p)*(z^q3)*KT3(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
%F A157704 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 A157704 Some PDGF3 (z;n) are:
%e A157704 PDGF3(z;n=3) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10
%e A157704 PDGF3(z;n=4) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10*(1-5*z)^13
%e A157704 The first few GFKT3's are:
%e A157704 GFKT3(z;p=0) = 1/(1-z)
%e A157704 GFKT3(z;p=1) = -(5*z+1)/(1-z)^4
%e A157704 GFKT3(z;p=2) = z*(32+186*z+132*z^2+10*z^3)/(1-z)^7
%e A157704 Some KT3(z,p) polynomials are:
%e A157704 KT3(z;p=2) = 32+186*z+132*z^2+10*z^3
%e A157704 KT3(z;p=3) = 56+2814*z+17834*z^2+27324*z^3+11364*z^4+1078*z^5+10*z^6
%p A157704 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-(k+1)*z)^(1+3*k), 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)); GFKT3(p):=sum((fk)*z^k, k=0..infinity); q3:=ldegree((numer(GFKT3(p)))): KT3(p):=sort((-1)^(p)*simplify((GFKT3(p)*(1-z)^(3*p+1))/z^q3),z, ascending);
%Y A157704 Originator sequence A156927.
%Y A157704 See A002414 for the z^1 coefficients and A157707 for the z^2 coefficients divided by 2.
%Y A157704 Row sums equal A064350 and those of A157705.
%Y A157704 Cf. A157702, A157703, A157705.
%K A157704 easy,nonn,tabf,uned
%O A157704 0,3
%A A157704 _Johannes W. Meijer_, Mar 07 2009