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 A147985 #18 Jul 01 2015 09:37:58 %S A147985 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, %T A147985 0,-220,0,83,0,-15,0,1,1,0,-31,0,413,0,-3141,0,15261,0,-50187,0, %U A147985 115410,0,-189036,0,222621,0,-189036,0,115410,0,-50187,0,15261,0,-3141,0,413,0 %N A147985 Coefficients of numerator polynomials S(n,x) associated with reciprocation. %C A147985 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). %C A147985 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.) %C A147985 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). %C A147985 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). %C A147985 5. If n>2, then S(n,1)=1 and S(n,2)=A127814(n). %C A147985 6. S(n,2^(1/2))=-1 for n>2 and S(n,2^(-1/2))=-2^(1-n) for n>1. %H A147985 Peter J. C. Moses, <a href="/A147985/b147985.txt">Rows n = 1..13 of irregular triangle, flattened</a> %H A147985 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Kimberling/kimberling56.html">Polynomials associated with reciprocation</a>, Journal of Integer Sequences 12 (2009, Article 09.3.4) 1-11. %F A147985 The basic idea is to iterate the reciprocation-difference mapping x/y -> x/y-y/x. %F A147985 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). %e A147985 S(1)=x %e A147985 S(2)=x^2-1=(x-1)(x+1) %e A147985 S(3)=x^4-3*x^2+1=(x^2+x-1)(x^2-x-1) %e A147985 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), %e A147985 so that, as an array, sequence begins with %e A147985 1 0 %e A147985 1 0 -1 %e A147985 1 0 -3 0 1 %e A147985 1 0 -7 0 13 0 -7 0 1 %t A147985 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 *) %Y A147985 Cf. A147986, A147987, A147988, A147989, A147990, A147991, A147992, A147993. %K A147985 sign,tabf %O A147985 1,8 %A A147985 _Clark Kimberling_, Nov 24 2008