OFFSET
1,10
COMMENTS
T(n)=S(1)*S(2)*...*S(n-1). The degree of S(n) in x is m=2^(n-1), so that the degree of T(n) is m-1. Write the zeros of T(n) as r(1)<r(2)<...<r(m-1) and the zeros of S(n) as z(1)<z(2)<...<z(m). Then z(1)<r(1)<z(2)<r(2)<...<r(m-1)<z(m); i.e., the zeros of T(n) intersperse the zeros of S(n).
LINKS
Clark Kimberling, Polynomials associated with reciprocation, Journal of Integer Sequences 12 (2009, Article 09.3.4) 1-11.
FORMULA
The basic idea is to iterate the reciprocation-difference mapping x/y -> x/y-y/x.
Let x be an indeterminate, S(1)=x, T(1)=1 and for n>1, define S(n)=S(n-1)^2-T(n-1)^2 and T(n)=S(n-1)*T(n-1), so that S(n)/T(n)=S(n-1)/T(n-1)-T(n-1)/S(n-1).
EXAMPLE
T(1) = 1
T(2) = x
T(3) = x^3-x
T(4) = x^7-4*x^5+4*x^3-x
so that, as an array, the sequence begins with:
1
1 0
1 0 -1 0
1 0 -4 0 4 0 -1 0
MATHEMATICA
s[1] = x; t[1] = 1; s[n_] := s[n] = s[n-1]^2 - t[n-1]^2; t[n_] := t[n] = s[n-1]*t[n-1]; row[n_] := CoefficientList[t[n], x] // Reverse; Table[row[n], {n, 7}] // Flatten (* Jean-François Alcover, Apr 22 2013 *)
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Clark Kimberling, Nov 24 2008
STATUS
approved