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%I #24 Feb 08 2022 07:51:38
%S 1,-2,1,0,-2,1,0,-12,4,1,0,-144,28,20,1,0,-2880,216,508,50,1,0,-86400,
%T -2592,17400,2548,98,1,0,-3628800,-449280,788688,153760,8568,168,1,0,
%U -203212800,-42405120,46032768,11269008,811648,23016,264,1,0,-14631321600,-4187635200,3372731136
%N Coefficients of the v=1 member of a family of certain orthogonal polynomials.
%C For v>=1 the orthogonal polynomials pt(n,v,x) have v integer zeros k*(k+1), k=1..v, for every n>=v and some other n-v zeros. The integer zeros are from 2*A000217.
%C The v-family pt(n,v,x) consists of characteristic polynomials of the tridiagonal M x M matrix Vt=Vt(M,v) with entries Vt_{m,n} given by 2*m*(v+1-m) if n=m, m=1,...,M; -m*(v+1-m) if n=m-1, m=2,...,M; -m*(v+1-m) if n=m+1, m=1..M-1 and 0 else. pt(n,v,x):=det(x*I_n-Vt(n,v)) with the n dimensional unit matrix I_n.
%C pt(n,v=1,x) has, for every n>=1, among its n zeros one for x=2. pt(1,1,x) has therefore only the integer zeros 2. det(Vt(1,1))=2.
%C The column sequences give [1,-2,0,0,0,...], A010790(n-1)*(-1)^(n-1), A130185, A130186 for m=0,1,2,3.
%C Coefficients of pt(n,v=1,x) (in the quoted Bruschi et al. paper {\tilde p}^{(\nu)}_n(x) of eqs. (20) and (24a),(24b)) in increasing powers of x.
%H M. Bruschi, F. Calogero and R. Droghei, <a href="http://dx.doi.org/10.1088/1751-8113/40/14/005">Proof of certain Diophantine conjectures and identification of remarkable classes of orthogonal polynomials</a>, J. Physics A, 40(2007), pp. 3815-3829.
%H Wolfdieter Lang, <a href="/A130182/a130182.txt">First ten rows and more.</a>
%F a(n,m) = [x^m]pt(n,1,x), n>=0, with the three term recurrence for orthogonal polynomial systems of the form pt(n,v,x) = (x + 2*n*(n-1-v))*pt(n-1,v,x) -(n-1)*n*(n-1-v)*(n-2-v)*pt(n-2,v,x), n>=1; pt(-1,v,x)=0 and pt(0,v,x)=1. Put v=1 here.
%F Recurrence: a(n,m) = a(n-1,m-1)+2*n*(n-2)*a(n-1,m) - (n-1)*n*(n-2)*(n-3)*a(n-2,m); a(n,m)=0 if n<m, a(-1,m):=0, a(n,-1)=1. From the pt(n,1,x) recurrence.
%e Triangle begins:
%e [1];
%e [-2,1];
%e [0,-2,1];
%e [0,-12,4,1];
%e [0,-144,28,20,1];
%e [0,-2880,216,508,50,1];
%e ...
%e Row n=5:[0,-2880,216,508,50,1]; pt(5,2,x)= x*(-2880+216*x+508*x^2+50*x^3+1*x^4)= x*(x-2)*(1440+612*x+52*x^2+x^3). pt(5,1,x) has the guaranteed integer zero x=2 (and also x=0 and some other three zeros).
%e Row n=1:[ -2,1]. pt(1,1,x)=-2+x with integer zero x=2.
%Y Cf. A129065 (a v=1 member of a similar family).
%Y Row sums A130183, unsigned row sums A130184.
%K sign,tabl,easy
%O 0,2
%A _Wolfdieter Lang_, Jun 01 2007
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