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.

A192342 Constant term of the reduction of n-th polynomial at A158983 by x^2->x+2.

Original entry on oeis.org

2, 7, 100, 28051, 2357659852, 16675673548656023155, 834234264904007920903714901139450715276, 2087840426219791385723375854976408025594408461778898567573217959566013061037427
Offset: 1

Views

Author

Clark Kimberling, Jun 28 2011

Keywords

Comments

For an introduction to reductions of polynomials by substitutions such as x^2->x+2, see A192232.

Examples

			The first three polynomials at A158983 and their reductions are as follows:
p0(x)=2+x -> 2+x
p1(x)=5+4x+x^2 -> 7+5x
p2(x)=26+40x+26x^2+8x^3+x^4 -> 100+95x.
From these, we read
A192342=(2,7,100,...) and A192343=(1,5,95,...)
		

Crossrefs

Programs

  • Mathematica
    q[x_] := x + 2;
    p[0, x_] := x + 2;
    p[n_, x_] := 1 + p[n - 1, x]^2 /; n > 0  (* polynomials defined at A158983 *)
    Table[Expand[p[n, x]], {n, 0, 4}]
    reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
       x^y_?OddQ -> x q[x]^((y - 1)/2)};
    t = Table[Last[Most[FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,9}]
    Table[Coefficient[Part[t, n], x, 0], {n, 1, 9}]
      (* A192342 *)
    Table[Coefficient[Part[t, n], x, 1], {n, 1, 9}]
      (* A192343 *)

A192343 Coefficient of x in the reduction of n-th polynomial at A158983 by x^2->x+2.

Original entry on oeis.org

1, 5, 95, 28025, 2357659175, 16675673548655564825, 834234264904007920903714900929384326375, 2087840426219791385723375854976408025594408461778898567529090071820106885049625
Offset: 1

Views

Author

Clark Kimberling, Jun 28 2011

Keywords

Comments

For an introduction to reductions of polynomials by substitutions such as x^2->x+2, see A192232.

Examples

			(See A192342.)
		

Crossrefs

Programs

A158985 Coefficients of polynomials (in descending powers of x) P(n,x) := 1 + P(n-1,x)^2, where P(1,x) = x + 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 8, 8, 5, 1, 8, 32, 80, 138, 168, 144, 80, 26, 1, 16, 128, 672, 2580, 7664, 18208, 35296, 56472, 74944, 82432, 74624, 54792, 31776, 13888, 4160, 677, 1, 32, 512, 5440, 43048, 269920, 1393728, 6082752, 22860480, 75010560, 217147904
Offset: 1

Views

Author

Clark Kimberling, Apr 02 2009

Keywords

Examples

			Row 1: 1 1 (from x + 1)
Row 2: 1 2 2 (from x^2 + 2*x + 2)
Row 3: 1 4 8 8 5
Row 4: 1 8 32 80 138 168 144 80 26
		

Crossrefs

Programs

  • PARI
    tabf(nn) = {my(P = x+1); print(Vec(P)); for (n=1, nn, P = 1 + P^2; print(Vec(P)););} \\ Michel Marcus, Jul 01 2015

Formula

From Peter Bala, Jul 01 2015: (Start)
P(n,x) = P(n,-2 - x) for n >= 2.
P(n+1,x)= P(n,(1 + x)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) - 1 is a zero of P(n+1,x).
Define a sequence of polynomials Q(n,x) by setting Q(1,x) = 1 + x^2 and Q(n,x) = Q(n-1, 1 + x^2) for n >= 2. Then P(n,x) = Q(n,sqrt(x)).
Q(n,x) = Q(k,Q(n-k,x)) for 1 <= k <= n-1; P(n,x) = P(k,P(n-k,x)^2) for 1 <= k <= n - 1.
n-th row sum = P(n,1) = A003095(n+1);
P(n,1) = P(n+1,0) = P(n+1,-2); P(n,1) = P(n,-3) for n >= 2.
P(n,2) = A062013(n). (End)

A158982 Coefficients of polynomials P(n,x):=-2+P(n-1,x)^2, where P(0,x)=x-2.

Original entry on oeis.org

1, -2, 1, -4, 2, 1, -8, 20, -16, 2, 1, -16, 104, -352, 660, -672, 336, -64, 2, 1, -32, 464, -4032, 23400, -95680, 283360, -615296, 980628, -1136960, 940576, -537472, 201552, -45696, 5440, -256, 2, 1, -64, 1952, -37760, 520144, -5430656, 44662464
Offset: 1

Views

Author

Clark Kimberling, Apr 02 2009

Keywords

Comments

The 2^n zeros of P(n,x) are 2+2*cos[(2k-1)Pi/(2^(n+1))], k=1,2,...,2^n.
P(n,x) = 2*T(2^(n+1),(1/2)x^(1/2)), where T(k,t) is the k-th Chebyshev polynomial of the first kind.

Examples

			Row 1: 1 -2 (from x-2)
Row 2: 1 -4 2 (from x^2-4x+2)
Row 3: 1 -8 20 -16 2
Row 4: 1 -16 104 -352 660 -672 336 -64 2
		

Crossrefs

Programs

  • PARI
    tabf(nn) = {p = x-2; print(Vec(p)); for (n=2, nn, p = -2 + p^2; print(Vec(p)););} \\ Michel Marcus, Mar 01 2016

Formula

P(n+1,x+2) = P(n,x^2) for n>=0.

A158984 Coefficients of polynomials (in descending powers of x) P(n,x) := -1 + P(n-1,x)^2, where P(1,x) = x - 1.

Original entry on oeis.org

1, -1, 1, -2, 0, 1, -4, 4, -1, 1, -8, 24, -32, 14, 8, -8, 0, 0, 1, -16, 112, -448, 1116, -1744, 1552, -384, -700, 736, -160, -128, 64, 0, 0, 0, -1, 1, -32, 480, -4480, 29112, -139552, 509600, -1441024, 3166616, -5345344, 6668992, -5473536, 1494624
Offset: 1

Views

Author

Clark Kimberling, Apr 02 2009

Keywords

Examples

			Row 1: 1 -1 (from x-1)
Row 2: 1 -2 0 (from x^2-2x)
Row 3: 1 -4 4 -1
Row 4: 1 -8 24 -32 14 8 -8 0 0
		

Crossrefs

Programs

  • PARI
    tabf(nn) = {p = x-1; print(Vec(p)); for (n=2, nn, p = -1 + p^2; print(Vec(p)););} \\ Michel Marcus, Mar 01 2016

Formula

From Peter Bala, Jul 01 2015: (Start)
P(n,x) = P(n,2 - x) for n >= 2.
P(n+1,x)= P(n,(x - 1)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) + 1 is a zero of P(n+1,x).
Define a sequence of polynomials Q(n,x) by setting Q(1,x) = -1 + x^2 and Q(n,x) = Q(n-1, -1 + x^2) for n >= 2. Then P(n,x) = Q(n,sqrt(x)).
Q(n,x) = Q(k,Q(n-k,x)) for 1 <= k <= n - 1; P(n,x) = P(k,P(n-k,x)^2) for 1 <= k <= n - 1.
P(n,x)^(2^k) divides P(n + 2*k,x) in Z[x] for k = 1,2,....
P(n,4) = A003096(n). (End)

A158986 Coefficients of polynomials Q(n,x):=-2+(1+Q(n-1,x))^2, where Q(1,x)=x-2.

Original entry on oeis.org

1, -2, 1, -2, -1, 1, -4, 4, 0, -2, 1, -8, 24, -32, 14, 8, -8, 0, -1, 1, -16, 112, -448, 1116, -1744, 1552, -384, -700, 736, -160, -128, 64, 0, 0, 0, -2, 1, -32, 480, -4480, 29112, -139552, 509600, -1441024, 3166616, -5345344, 6668992, -5473536, 1494624, 3005056, -4820608
Offset: 1

Views

Author

Clark Kimberling, Apr 02 2009

Keywords

Comments

Let P(n,x) be the n-th polynomial at A158984. Then Q(n,x)=P(n-1,x)-1 is a factor of P(n,x).

Examples

			Row 1: 1 -2 (from x-2)
Row 2: 1 -2 -1 (from x^2-2x-1)
Row 3: 1 -4 4 0 -2
Row 4: 1 -8 24 -32 14 8 -8 0 -1
		

Crossrefs

Programs

  • PARI
    tabf(nn) = {p = x-2; print(Vec(p)); for (n=2, nn, p = -2 + (p+1)^2; print(Vec(p)););} \\ Michel Marcus, Mar 01 2016
Showing 1-6 of 6 results.