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%I #13 Apr 15 2021 21:46:43
%S 1,1,1,1,-6,1,7,-79,119,126,-270,1,28,-515,1654,8689,-65864,142371,
%T -82242,-99090,113400,1,86,-2255,5784,300930,-3904584,20663714,
%U -41517272,-80232259,657717054
%N FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle.
%C The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below.
%C An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely.
%C A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.
%F G.f.: GF1(z;RHCnr) := FP1(z;RHCnr)/product((1-(2*m-1)*z)^(RHCnr+1-m),m=1..RHCnr)
%F Row sums (n) = (-1)^(1+(n+1)*(n+2)/2)*A098695(n).
%e The first few rows of the "triangle" of the coefficients of the FP1 polynomials.
%e In the columns the coefficients of the powers of z^m, m=0,1,2,... , appear.
%e [1]
%e [1]
%e [1, 1, -6]
%e [1, 7, -79, 119, 126, -270]
%e [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
%e Matrix of the coefficients of the FP1 polynomials. The coefficients in the columns of this matrix are the powers of z^m, m=0,1,2,.. .
%e [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
%e [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
%e [1, 1, -6, 0 ,0, 0, 0, 0, 0, 0]
%e [1, 7, -79, 119, 126, -270, 0, 0, 0, 0]
%e [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
%e The first few FP1 polynomials are:
%e FP1(z; RHCnr=1) = 1
%e FP1(z; RHCnr=2) = 1
%e FP1(z; RHCnr =3) = 1+z-6*z^2
%e Some GF1(z;RHCnr) are:
%e GF1(z;RHCnr= 3) = (1+z-6*z^2)/((1-5*z)*(1-3*z)^2*(1-z)^3)
%e GF1(z;RHCnr= 4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)
%p RHCnr:=4: if RHCnr=1 then RHCmax :=1; else RHCmax:=(RHCnr-1)*(RHCnr)/2 end if: RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr+1); for nx from 0 to RHCmax do num:=sort(sum(A[t]*z^t, t=0..RHCmax)); nom:=Product((1-(2*u-1)*z)^(RHCnr-u+1),u=1..RHCnr): RHCa:= series(num/nom,z,nx+1); y:=coeff(RHCa,z,nx)-A[nx]; x:=RHC[nx+1]; A[nx]:=x-y; end do: FP1[RHCnr]:=sort(num,z, ascending); GenFun[RHCnr] :=FP1[RHCnr]/product((1-(2*m-1)*z)^(RHCnr-m+1),m=1..RHCnr);
%Y Cf. A156920, A156925, A156927, A156933.
%Y For the first few GF1's see A000340, A156922, A156923, A156924.
%Y The number of FP1 terms follow the triangular numbers A000217, with quite surprisingly one exception here a(0)=1.
%Y Abs(Row sums (n)) = A098695(n).
%Y For the polynomials in the denominators of the GF1(z;RHCnr) see A157702.
%K easy,sign,tabf,uned
%O 0,5
%A _Johannes W. Meijer_, Feb 20 2009
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