cp's OEIS Frontend

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.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Nov 24 2008

Keywords

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
		

Crossrefs

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).

A147986 Coefficients of denominator polynomials T(n,x) associated with reciprocation.

Original entry on oeis.org

1, 1, 0, 1, 0, -1, 0, 1, 0, -4, 0, 4, 0, -1, 0, 1, 0, -11, 0, 45, 0, -88, 0, 88, 0, -45, 0, 11, 0, -1, 0, 1, 0, -26, 0, 293, 0, -1896, 0, 7866, 0, -22122, 0, 43488, 0, -60753, 0, 60753, 0, -43488, 0, 22122, 0, -7866, 0, 1896, 0, -293, 0, 26, 0, -1, 0, 1, 0, -57, 0, 1512, 0
Offset: 1

Views

Author

Clark Kimberling, Nov 24 2008

Keywords

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)

Examples

			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
		

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[t[n], x] // Reverse; Table[row[n], {n, 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).

A147990 Array A147985 (Polynomial coefficients) with zeros deleted.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -7, 13, -7, 1, 1, -15, 83, -220, 303, -220, 83, -15, 1, 1, -31, 413, -3141, 15261, -50187, 115410, -189036, 222621, -189036, 115410, -50187, 15261, -3141, 413, -31, 1, 1, -63, 1839, -33150, 414861, -3841195, 27378213, -154299168
Offset: 1

Author

Clark Kimberling, Nov 25 2008

Keywords

Examples

			s(1)=x
s(2)=S(2,y)=x-1
s(3)=S(3,y)=x^2-3*x+1
s(4)=S(4,y)=x^4-7*x^3+13*x^2-7*x+1
so that as an array A147990 begins with
1
1 -1
1 -3 1
1 -7 13 -7 1
		

Formula

Let s(1)=x and for n>=2, let s(n)=s(n,x)=S(n,y), where y=x^(1/2) and S(n,x)
is as at A147985. Then A147990 gives the coefficients of the polynomials s(n).

A147987 Coefficients of numerator polynomials P(n,x) associated with reciprocation.

Original entry on oeis.org

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, 31, 0, 1, 1, 0, 63
Offset: 1

Author

Clark Kimberling, Nov 24 2008

Keywords

Comments

1. P(n,1)=A073833(n) for n>=1; P(n,2)=A073833(n+1) for n>=0.
2. P(n)=P(n-1)^2+P(n-1)*P(n-2)^2-P(n-2)^4 for n>=3.
3. For n>=3, P(n)=P(n,x)=S(n,i*x), where S(n) is the polynomial at A147985.
Thus all the zeros of P(n,x), for n>=2, are nonreal.

Examples

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

Programs

  • Mathematica
    p[1] = x; q[1] = 1; p[n_] := p[n] = p[n-1]^2 + q[n-1]^2; q[n_] := q[n] = p[n-1]*q[n-1]; row[n_] := CoefficientList[p[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-sum mapping x/y -> x/y+y/x.
Let x be an indeterminate, P(1)=x, Q(1)=1 and for n>1, define P(n)=P(n-1)^2+Q(n-1)^2 and Q(n)=P(n-1)*Q(n-1), so that P(n)/Q(n)=P(n-1)/Q(n-1)-Q(n-1)/P(n-1).

A147988 Coefficients of denominator polynomials Q(n,x) associated with reciprocation.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 0, 1, 0, 4, 0, 4, 0, 1, 0, 1, 0, 11, 0, 45, 0, 88, 0, 88, 0, 45, 0, 11, 0, 1, 0, 1, 0, 26, 0, 293, 0, 1896, 0, 7866, 0, 22122, 0, 43488, 0, 60753, 0, 60753, 0, 43488, 0, 22122, 0, 7866, 0, 1896, 0, 293, 0, 26, 0, 1, 0, 1, 0, 57, 0, 1512, 0, 24858, 0, 284578, 0
Offset: 1

Author

Clark Kimberling, Nov 24 2008

Keywords

Comments

1. Q(n,1)=A073834(n) for n>=1.
2. For n>=3, Q(n)=Q(n,x)=i*T(n,i*x), where T(n) is the polynomial at A147986.
Thus all the zeros of Q(n,x), for n>=2, are nonreal.

Examples

			Q(1) = 1
Q(2) = x
Q(3) = x^3+x
Q(4) = x^7+4*x^5+4*x^3+1
so that, as an array, the sequence begins with:
1
1 0
1 0 1 0
1 0 4 0 4 0 1
		

Formula

The basic idea is to iterate the reciprocation-sum mapping x/y -> x/y+y/x.
Let x be an indeterminate, P(1)=x, Q(1)=1 and for n>1, define P(n)=P(n-1)^2+Q(n-1)^2 and Q(n)=P(n-1)*Q(n-1), so that P(n)/Q(n)=P(n-1)/Q(n-1)-Q(n-1)/P(n-1).
Showing 1-5 of 5 results.