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 1-6 of 6 results.

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

Views

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

Links

Crossrefs

Cf. A147985, A147986, A147987, A147988, A147989, A147991, A147992, A147993.

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).

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

Links

Crossrefs

Cf. A147985, 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[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).

A147989 Coefficients of factor polynomials U(n,x) associated with reciprocation.

Original entry on oeis.org

1, 1, -1, 1, 1, -3, -1, 1, 1, 1, -7, -4, 13, 4, -7, -1, 1, 1, 1, -15, -11, 83, 45, -220, -88, 303, 88, -220, -45, 83, 11, -15, -1, 1, 1, 1, -31, -26, 413, 293, -3141, -1896, 15261, 7866, -50187, -22122, 115410, 43488, -189036, -60753, 222621, 60753, -189036
Offset: 1

Views

Author

Clark Kimberling, Nov 25 2008

Keywords

Comments

The zeros of U(n,x) and U(n,-x) are the zeros of S(n,x) at A147985.

Examples

			U(3) = x^2+x-1;
U(4) = x^4+x^3-3*x^2-x+1;
U(5) = x^8+x^7-7*x^6-4*x^5+13*x^4+4*x^3-7*x^2-x+1;
so that, as an array, the sequence begins with:
1 1 -1
1 1 -3 -1 1
1 1 -7 -4 13 4 -7 -1 1

Links

Crossrefs

Cf. A147985, A147986, A147987, A147988, A147990, A147991, A147992, A147993.

Programs

  • Maple
    U[3]:= x^2+x-1:
U[4]:= x^4+x^3-3*x^2-x+1:
for n from 5 to 10 do
  U[n]:= normal(U[n-1]*M(U[n-1]) + x*(x^2-1)*mul(U[i]*M(U[i]),i=3..n-2));
od:
seq(seq(coeff(U[m],x,j),j=degree(U[m])..0,-1),m=3..10); # Robert Israel, Jun 30 2015
  • Mathematica
    U[3, x_] = x^2 + x - 1;
U[4, x_] = x^4 + x^3 - 3 x^2 - x + 1;
U[n_, x_] := U[n, x] = U[n-1, x] U[n-1, -x] + x (x^2 - 1) Product[U[k, x] U[k, -x], {k, 3, n-2}];
Table[CoefficientList[U[n, x], x] // Reverse, {n, 3, 7}] // Flatten (* Jean-François Alcover, Mar 25 2019 *)

Formula

For n>=5, U(n)=U(n,x)=U(n-1,x)*U(n-1,-x)+x*(x^2-1)*U(3,x)*U(3,-x)*U(4,x)*U(4,-x)*...*U(n-2,x)*U(n-2,-x), where U(3)=x^2+x-1, U(4)=x^4+x^3-3*x^2-x+1.

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

Views

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

Links

Crossrefs

Cf. A147985, A147986, A147988, A147989, A147990, A147991, A147992, A147993.

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

Views

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

Links

Crossrefs

Cf. A147985, A147986, A147987, A147989, A147990, A147991, A147992, A147993.

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).

A154305 Coefficients of polynomials H(n,x) associated with squares of polynomials S(n,x).

Original entry on oeis.org

1, 0, 1, 1, 0, -6, 0, 1, 1, 0, 20, 0, -26, 0, 20, 0, 1, 1, 0, -88, 0, 92, 0, -872, 0, 1990, 0, -872, 0, 92, 0, -88, 0, 1, 1, 0, 336, 0, -3336, 0, 6961, 0, -77796, 0, -647088, 0, 2618568, 0, -3600784, 0, 3346502, 0, -3600784, 0, 2618568, 0, -647088, 0, -77796, 0
Offset: 1

Views

Author

Clark Kimberling, Jan 06 2009

Keywords

Comments

Define S(1)=S(1,x)=x and T(1)=T(1,x)=1; for n>=1, define S(n+1)=[S(n)]^2-[T(n)]^2 and T(n+1)=c*S(n)*T(n). The sole value of c for which S(n) is the square of a polynomial for all n>=3 is 2i, and [H(n,x)]^2 = S(n,x).

Examples

			H(3,x)=x^2+1 and S(3,x)=(x^2+1)^2.
H(4,x)=x^4-6*x^2+1
H(6,x)=x^8+20*x^6-26*x^4+20*x^2+1.
First three rows:
1 0 1
1 0 -6 0 1
1 0 20 0 -26 0 20 0 1.

Links

Crossrefs

Cf. A147985, A147986.

Formula

H(3,x)=x^2+1 and H(n+1,x)=[(2*i*x)^p]*H(n,i/(2*x)-ix/2) for n>=3, where p=2^n-2 and i=sqrt(-1).
H(n,x)=2*H(n-2,x)^4-H(n-1,x)^2. [Clark Kimberling, Mar 19 2009]
Showing 1-6 of 6 results.

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