A147985 Coefficients of numerator polynomials S(n,x) associated with reciprocation.

1, 0, 1, 0, -1, 1, 0, -3, 0, 1, 1, 0, -7, 0, 13, 0, -7, 0, 1, 1, 0, -15, 0, 83, 0, -220, 0, 303, 0, -220, 0, 83, 0, -15, 0, 1, 1, 0, -31, 0, 413, 0, -3141, 0, 15261, 0, -50187, 0, 115410, 0, -189036, 0, 222621, 0, -189036, 0, 115410, 0, -50187, 0, 15261, 0, -3141, 0, 413, 0

Offset: 1

Author

Clark Kimberling, Nov 24 2008

Keywords

• sign
• tabf

Comments

1. S(n)=U(n-1)V(n-1) where U(n-1)=S(n-1)+S(1)*S(2)*...*S(n-2) and V(n-1)=S(n-1)-S(1)*S(2)*...*S(n-2), for n>=2. If U(n) and V(n) are written as polynomials U(n,x) and V(n,x), then V(n,x)=U(n,-x). See A147989 for coefficients of U(n).

2. S(n)=S(n-1)^2+S(n-1)*S(n-2)^2-S(n-2)^4 for n>2. (The Gorskov-Wirsting polynomials also have this recurrence; see H. L. Montgomery, Ten Lectures on the Interface between Analytic Number Theory and Harmonic Analysis, CBMS Regional Conference Series in Mathematics, 84, AMS, pp. 183-190.)

3. For n>0, the 2^(n-1) zeros of S(n) are real. If r is a zero of S(n), then -r and 1/r are zeros of S(n).

4. If r is a zero of S(n), then the numbers z satisfying r=z-1/z and r=z+1/z are zeros of S(n+1).

5. If n>2, then S(n,1)=1 and S(n,2)=A127814(n).

6. S(n,2^(1/2))=-1 for n>2 and S(n,2^(-1/2))=-2^(1-n) for n>1.

Examples

			S(1)=x
S(2)=x^2-1=(x-1)(x+1)
S(3)=x^4-3*x^2+1=(x^2+x-1)(x^2-x-1)
S(4)=x^8-7*x^6+13*x^4-7*x^2+1=(x^4+x^3-3*x^2-x+1)(x^4-x^3-3*x^2+x+1),
so that, as an array, sequence begins with
1 0
1 0 -1
1 0 -3 0 1
1 0 -7 0 13 0 -7 0 1
		

Links

• Peter J. C. Moses, Rows n = 1..13 of irregular triangle, flattened
• Clark Kimberling, Polynomials associated with reciprocation, Journal of Integer Sequences 12 (2009, Article 09.3.4) 1-11.

Crossrefs

Cf. A147986, A147987, A147988, A147989, A147990, A147991, A147992, A147993.

Programs

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[s[n], x] // Reverse; Table[row[n], {n, 1, 7}] // Flatten (* Jean-François Alcover, Apr 22 2013 *)

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).