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.

A192340 Constant term of the reduction of n-th polynomial at A158985 by x^2->x+1.

Original entry on oeis.org

1, 3, 19, 1091, 4270307, 65975813893475, 15748607358316275150858234851, 897339846665475127909937786392825941994036757434025817827, 2913308988276889310145046342161059349226587591969604604068795694857825566722967409631885309325418272374141705507555
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+1, see A192232.

Examples

			The first three polynomials at A158985 and their reductions are as follows:
p0(x)=1+x -> 1+x
p1(x)=2+2x+x^2 -> 3+3x
p2(x)=5+8x+8x^2+4x^3+x^4 -> 19+27x.
From these, we read
A192340=(1,3,19,...) and A192341=(1,3,27,...)

Crossrefs

Cf. A192232, A192341.

Programs

  • Mathematica
    q[x_] := x + 1;
p[0, x_] := x + 1;
p[n_, x_] := 1 + p[n - 1, x]^2 /; n > 0  (* polynomials defined at A158985 *)
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}]
(* A192340 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 9}]
(* A192341 *)

A192341 Coefficient of x in the reduction of n-th polynomial at A158985 by x^2->x+1.

Original entry on oeis.org

1, 3, 27, 1755, 6909435, 106751109312315, 25481781981232427172041948475, 1451926371364357751230060437820316313546007023604937430075
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+1, see A192232.

Examples

			(See A192340.)

Crossrefs

Cf. A192232, A192340.

Programs

A158983 Coefficients of polynomials (in descending powers of x) P(n,x) := 2 + P(n-1,x)^2, where P(1,x) = x + 2.

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 28, 48, 38, 1, 16, 120, 544, 1628, 3296, 4432, 3648, 1446, 1, 32, 496, 4928, 35064, 189248, 800992, 2711424, 7419740, 16475584, 29610272, 42666880, 48398416, 41867904, 26125248, 10550016, 2090918, 1, 64, 2016, 41600, 631536
Offset: 1

Views

Author

Clark Kimberling, Apr 02 2009

Keywords

Examples

			Row 1: 1 2 (from x+2)
Row 2: 1 4 6 (from x^2+4x+6)
Row 3: 1 8 28 48 38
Row 4: 1 16 120 544 1628 3296 4432 3648 1446

Links

Crossrefs

Cf. A158982, A158984, A158985, A158986. A102847 (row sums).

Formula

From Peter Bala, Jul 01 2015: (Start)
P(n,x) = P(n,-4 - x) for n >= 2.
P(n+1,x)= P(n,(2 + x)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) - 2 is a zero of P(n+1,x).
Define a sequence of polynomials Q(n,x) by setting Q(1,x) = 2 + x^2 and Q(n,x) = Q(n-1, 2 + 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) = A102847(n);
P(n,1) = P(n+1,-1) = P(n+1,-3); P(n,1) = P(n,-5) for n >= 2.
(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

Links

Crossrefs

Cf. A084534, A158983, A158984, A158985, A158986.

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

Links

Crossrefs

Cf. A158982, A158983, A158985, A158986, A003096.

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

Links

Crossrefs

Cf. A158982, A158983, A158984, A158985.

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.

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