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.

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A266800 Coefficient of x in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-8, -12, -98, -636, -4424, -30138, -207032, -1417788, -9720866, -66619404, -456638168, -3129787002, -21452029928, -147034005996, -1007787102434, -6907472856348, -47344530365672, -324504220137018, -2224185061818776, -15244791078484764, -104489352838678178
Offset: 0

Views

Author

Clark Kimberling, Jan 09 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(3),1,1,1,...] has p(0,x)=1-8x-7x^2+2x^3+x^4, so a(0) = -8;
[1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = -12;
[1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = -98.

Links

Crossrefs

Cf. A265762, A266799, A266801, A266802.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266799 *)
Coefficient[t, x, 1];  (* A266800 *)
Coefficient[t, x, 2];  (* A266801 *)
Coefficient[t, x, 3];  (* A266802 *)
Coefficient[t, x, 4];  (* A266799 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: -((2 (-4 + 14 x + 41 x^2 - 43 x^3 + 3 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5)).

A266801 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-7, 23, 65, 653, 3935, 28373, 190793, 1317335, 9003953, 61779965, 423273503, 2901611813, 19886759705, 136308977303, 934267517345, 6403586065133, 43890776239583, 300832001287925, 2061932830446953, 14132698865151575, 96866956468010513, 663936003630421853
Offset: 0

Views

Author

Clark Kimberling, Jan 09 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(3),1,1,1,...] has p(0,x)=1-8x-7x^2+2x^3+x^4, so a(0) = -7;
[1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = 23;
[1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = 65.

Links

Crossrefs

Cf. A265762, A266799, A266800, A266802.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266799 *)
Coefficient[t, x, 1];  (* A266800 *)
Coefficient[t, x, 2];  (* A266801 *)
Coefficient[t, x, 3];  (* A266802 *)
Coefficient[t, x, 4];  (* A266799 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (7 - 58 x - 55 x^2 + 122 x^3 - 5 x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

A266802 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

2, -12, -16, -294, -1552, -11868, -78142, -543996, -3706624, -25463142, -174376288, -1195587372, -8193644926, -56162781804, -384938354032, -2638425262758, -18083987259952, -123949619666556, -849562999302334, -5822992294650972, -39911380656754528
Offset: 0

Views

Author

Clark Kimberling, Jan 09 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(3),1,1,1,...] has p(0,x) = 1 - 8 x - 7 x^2 + 2 x^3 + x^4, so a(0) = 2;
[1,sqrt(3),1,1,1,...] has p(1,x) = 1 - 12 x + 23 x^2 - 12 x^3 + x^4, so a(1) = -12;
[1,1,sqrt(3),1,1,1...] has p(2,x) = 49 - 98 x + 65 x^2 - 16 x^3 + x^4, so a(2) = -16.

Links

Crossrefs

Cf. A265762, A266299, A266800, A266801.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[3]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266799 *)
Coefficient[t, x, 1];  (* A266800 *)
Coefficient[t, x, 2];  (* A266801 *)
Coefficient[t, x, 3];  (* A266802 *)
Coefficient[t, x, 4];  (* A266799 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (2 (-1 + 11 x - 7 x^2 + 2 x^3 + 6 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

A266804 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.

Original entry on oeis.org

19, 19, 361, 1795, 14011, 91489, 638899, 4348051, 29883145, 204609571, 1402971259, 9614651329, 65903614291, 451700107795, 3096024736681, 21220400800579, 145446970016059, 996907894114081, 6832909585226995, 46833455808339091, 321001289959109449
Offset: 0

Views

Author

Clark Kimberling, Jan 09 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = 19;
[1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = 19;
[1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = 361.

Links

Crossrefs

Cf. A265762, A266805, A266806, A266807.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266804 *)
Coefficient[t, x, 1];  (* A266805 *)
Coefficient[t, x, 2];  (* A266806 *)
Coefficient[t, x, 3];  (* A266807 *)
Coefficient[t, x, 4];  (* A266804 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (-19 + 76 x + 19 x^2 + 10 x^3 - x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

A266805 Coefficient of x in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-14, -90, -722, -4830, -33554, -228954, -1572110, -10768122, -73825010, -505954014, -3467991794, -23769625530, -162920337422, -1116670248090, -7653777913874, -52459758093534, -359564573392850, -2464492138756122, -16891880703949070, -115778671987640634
Offset: 0

Views

Author

Clark Kimberling, Jan 09 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(6),1,1,1,...] has p(0,x) = 19-14x-13x^2+2x^3+x^4, so a(0) = -14;
[1,sqrt(6),1,1,1,...] has p(1,x) = 19-90x+143x^2-90x^3+19x^4, so a(1) = -90;
[1,1,sqrt(6),1,1,1...] has p(2,x) = 361-722x+527x^2-166x^3+19x^4, so a(2) = -722.

Links

Crossrefs

Cf. A265762, A266804, A266806, A266807.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266804 *)
Coefficient[t, x, 1];  (* A266805 *)
Coefficient[t, x, 2];  (* A266806 *)
Coefficient[t, x, 3];  (* A266807 *)
Coefficient[t, x, 4];  (* A266804 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: -((2 (-7 - 10 x - 31 x^2 - 40 x^3 + 3 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5)).

A266806 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones. S.

Original entry on oeis.org

-13, 143, 527, 4859, 30119, 214847, 1450643, 10000367, 68393039, 469166939, 3214686407, 22036489343, 151033273907, 1035215971919, 7095427362959, 48632909524667, 333334588608743, 2284710128883647, 15659633909836499, 107332733533045679, 735669484346002127
Offset: 0

Views

Author

Clark Kimberling, Jan 10 2016

Keywords

Comments

ee A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = -13;
[1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = 143;
[1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = 527.

Links

Crossrefs

Cf. A265762, A266804, A266805, A266807.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266804 *)
Coefficient[t, x, 1];  (* A266805 *)
Coefficient[t, x, 2];  (* A266806 *)
Coefficient[t, x, 3];  (* A266807 *)
Coefficient[t, x, 4];  (* A266804 *)
  • PARI
    Vec((13-208*x-7*x^2+116*x^3+x^4)/(-1+5*x+15*x^2-15*x^3-5*x^4+x^5) + O(x^200)) \\ Altug Alkan, Jan 10 2015

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (13 - 208 x - 7 x^2 + 116 x^3 + x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

A266807 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(6),1,1,...], where 1^n means n ones.

Original entry on oeis.org

2, -90, -166, -2166, -12010, -89598, -594910, -4127706, -28160326, -193357590, -1324392298, -9079876830, -62228230846, -426534794586, -2923470679270, -20037876860598, -137341361295850, -941352453457086, -6452123715212446, -44223519044857050
Offset: 0

Views

Author

Clark Kimberling, Jan 10 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[sqrt(6),1,1,1,...] has p(0,x)=19-14x-13x^2+2x^3+x^4, so a(0) = 2;
[1,sqrt(6),1,1,1,...] has p(1,x)=19-90x+143x^2-90x^3+19x^4, so a(1) = -90;
[1,1,sqrt(6),1,1,1...] has p(2,x)=361-722x+527x^2-166x^3+19x^4, so a(2) = -166. ~

Links

Crossrefs

Cf. A265762, A266804, A266805, A266806.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[6]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266804 *)
Coefficient[t, x, 1];  (* A266805 *)
Coefficient[t, x, 2];  (* A266806 *)
Coefficient[t, x, 3];  (* A266807 *)
Coefficient[t, x, 4];  (* A266804 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (2 (-1 + 50 x - 127 x^2 - 22 x^3 + 15 x^4))/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).

A265802 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,4,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

1, 11, 19, 59, 145, 389, 1009, 2651, 6931, 18155, 47521, 124421, 325729, 852779, 2232595, 5845019, 15302449, 40062341, 104884561, 274591355, 718889491, 1882077131, 4927341889, 12899948549, 33772503745, 88417562699, 231480184339, 606022990331, 1586588786641
Offset: 0

Views

Author

Clark Kimberling, Jan 04 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[4,1,1,1,1,...] = (7 + sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = 1;
[1,4,1,1,1,...] = (29 - sqrt(5))/22 has p(1,x) = 19 - 29 x + 11 x^2, so a(1) = 11;
[1,1,4,1,1,...] = (67 + sqrt(5))/38 has p(2,x) = 59 - 67 x + 19 x^2, so a(2) = 19.

Links

Crossrefs

Cf. A000032, A000045, A265762, A265803.

Programs

  • GAP
    List([0..30], n-> 6*Fibonacci(n+1)^2 - 5*(-1)^n); # G. C. Greubel, Dec 11 2019
  • Magma
    I:=[1,11,19,59]; [n le 4 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 06 2016
  • Maple
    with(combinat); seq(6*fibonacci(n+1)^2 - 5*(-1)^n, n=0..30); # G. C. Greubel, Dec 11 2019
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {4}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A265802 *)
Coefficient[t, x, 1] (* A265803 *)
Coefficient[t, x, 2] (* A236802 *)
Join[{1}, LinearRecurrence[{2, 2, -1}, {11, 19, 59}, 30]] (* Vincenzo Librandi, Jan 06 2016 *)
Table[6*Fibonacci[n+1]^2 - 5*(-1)^n, {n,0,30}] (* G. C. Greubel, Dec 11 2019 *)
  • PARI
    Vec((1+9*x-5*x^2)/(1-2*x-2*x^2+x^3) + O(x^30)) \\ Altug Alkan, Jan 04 2016
  • PARI
    vector(31, n, 6*fibonacci(n)^2 + 5*(-1)^n) \\ G. C. Greubel, Dec 11 2019
  • Sage
    [6*fibonacci(n+1)^2 - 5*(-1)^n for n in (0..30)] # G. C. Greubel, Dec 11 2019

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n>3.
G.f.: (1 + 9*x - 5*x^2)/(1 - 2*x - 2*x^2 + x^3).
a(n) = (2^(-n)*(-13*(-2)^n + 3*(3-sqrt(5))^(1+n) + 3*(3+sqrt(5))^(1+n)))/5. - Colin Barker, Oct 20 2016
From Klaus Purath, Oct 28 2019: (Start)
(a(n-3) - a(n-2) - a(n-1) + a(n))/6 = Fibonacci(2*n-1).
(a(n-5) + a(n))/30 = Fibonacci(2*n-3).
(a(n) - a(n-4))/18 = Fibonacci(2*n-2). (End)
E.g.f.: (1/5)*exp(-x)*(-13 + exp(-(1/2)*(-5 + sqrt(5))*x)*(9 - 3*sqrt(5) + 3*(3 + sqrt(5))*exp(sqrt(5)*x))). - Stefano Spezia, Dec 09 2019
a(n) = 6*Fibonacci(n+1)^2 - 5*(-1)^n = (6*Lucas(2*n+2) - 13*(-1)^n)/5. - G. C. Greubel, Dec 11 2019

A265803 Coefficient of x in minimal polynomial of the continued fraction [1^n,4,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

-7, -29, -67, -185, -475, -1253, -3271, -8573, -22435, -58745, -153787, -402629, -1054087, -2759645, -7224835, -18914873, -49519771, -129644453, -339413575, -888596285, -2326375267, -6090529529, -15945213307, -41745110405, -109290117895, -286125243293
Offset: 0

Views

Author

Clark Kimberling, Jan 04 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[4,1,1,1,1,...] = (7 + sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = 1;
[1,4,1,1,1,...] = (29 - sqrt(5))/22 has p(1,x) = 19 - 29 x + 11 x^2, so a(1) = 11;
[1,1,4,1,1,...] = (67 + sqrt(5))/38 has p(2,x) = 59 - 67 x + 19 x^2, so a(2) = 19.

Links

Crossrefs

Cf. A000032, A000045, A265762, A265802.

Programs

  • GAP
    List([0..30], n-> (13*(-1)^n -12*Lucas(1,-1,2*n+3)[2])/5 ); # G. C. Greubel, Dec 12 2019
  • Magma
    I:=[-7,-29,-67]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 06 2016
  • Maple
    with(combinat); f:=fibonacci; seq( 5*(-1)^n - 12*f(n+1)*f(n+2), n=0..30); # G. C. Greubel, Dec 12 2019
  • Mathematica
    u[n_]:= Table[1, {k,n}]; t[n_]:= Join[u[n], {4}, {{1}}];
f[n_]:= FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A265802 *)
Coefficient[t, x, 1] (* A265803 *)
Coefficient[t, x, 2] (* A236802 *)
LinearRecurrence[{2,2,-1}, {-7,-29,-67}, 30] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    Vec((-7-15*x+5*x^2)/(1-2*x-2*x^2+x^3) + O(x^30)) \\ Altug Alkan, Jan 04 2016
  • PARI
    vector(31, n, f=fibonacci; -(5*(-1)^n + 12*f(n)*f(n+1)) ) \\ G. C. Greubel, Dec 12 2019
  • Sage
    [(13*(-1)^n -12*lucas_number2(2*n+3,1,-1))/5 for n in (0..30)] # G. C. Greubel, Dec 12 2019

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (-1)*(7 + 15*x - 5*x^2)/(1 - 2*x - 2*x^2 + x^3).
a(n) = (2^(-n)*(13*(-2)^n + 12*(3-sqrt(5))^n*(-2+sqrt(5)) - 12*(2+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Oct 20 2016
a(n) = (13*(-1)^n - 12*Lucas(2*n+3))/5 = 5*(-1)^n - 12*F(n+1)*F(n+2), F=Fibonacci. - G. C. Greubel, Dec 12 2019

A265804 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,5,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

1, 19, 29, 95, 229, 619, 1601, 4211, 11005, 28831, 75461, 197579, 517249, 1354195, 3545309, 9281759, 24299941, 63618091, 166554305, 436044851, 1141580221, 2988695839, 7824507269, 20484825995, 53629970689, 140405086099, 367585287581, 962350776671
Offset: 0

Views

Author

Clark Kimberling, Jan 05 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[5,1,1,1,1,...] = (9+sqrt(5))/2 has p(0,x) = 19 - 9 x + x^2, so a(0) = 1;
[1,5,1,1,1,...] = (47-sqrt(5))/38 has p(1,x) = 29 - 47 x + 19 x^2, so a(1) = 19;
[1,1,5,1,1,...] = (105+sqrt(5))/58 has p(2,x) = 5 - 105 x + 29 x^2, so a(2) = 29.

Links

Crossrefs

Cf. A265762, A265805.

Programs

  • Magma
    I:=[1,19,29]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 06 2016
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {5}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A265804 *)
Coefficient[t, x, 1] (* A265805 *)
Coefficient[t, x, 2] (* A236804 *)
LinearRecurrence[{2, 2, -1}, {1, 19, 29}, 30] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    Vec((1+17*x-11*x^2)/(1-2*x-2*x^2+x^3) + O(x^100)) \\ Altug Alkan, Jan 07 2016

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1 + 17 x - 11 x^2)/(1 - 2 x - 2 x^2 + x^3).
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