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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A265805 Coefficient of x in minimal polynomial of the continued fraction [1^n,5,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

-9, -47, -105, -295, -753, -1991, -5193, -13615, -35625, -93287, -244209, -639367, -1673865, -4382255, -11472873, -30036391, -78636273, -205872455, -538981065, -1411070767, -3694231209, -9671622887, -25320637425, -66290289415, -173550230793, -454360402991
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:=[-9,-47,-105]; [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}, {-9, -47, -105}, 30] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    Vec((-9-29*x+7*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.: (-9 - 29 x + 7 x^2)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(27*(-2)^n + 4*(3-sqrt(5))^n*(-9+5*sqrt(5)) - 4*(3+sqrt(5))^n*(9+5*sqrt(5))))/5. - Colin Barker, Oct 20 2016

A266699 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,1/2,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

4, 5, 1, 16, 29, 89, 220, 589, 1529, 4016, 10501, 27505, 71996, 188501, 493489, 1291984, 3382445, 8855369, 23183644, 60695581, 158903081, 416013680, 1089137941, 2851400161, 7465062524, 19543787429, 51166299745, 133955111824, 350699035709, 918141995321
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:
[1/2,1,1,1,1,...] = sqrt(5)/2 has p(0,x) = -5 + 4*x^2, so a(0) = 4;
[1,1/2,1,1,1,...] = (5 + 2*sqrt(5))/5 has p(1,x) = 1 - 10*x + 5*x^2, so a(1) = 5;
[1,1,1/2,1,1,...] = 6 - 2*sqrt(5) has p(2,x) = 16 - 12*x + x^2, so a(2) = 1.

Links

Crossrefs

Cf. A265762, A266700.

Programs

  • Magma
    I:=[4,5,1,16]; [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
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/2}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266699 *)
Coefficient[t, x, 1] (* A266700 *)
Coefficient[t, x, 2] (* A266699 *)
Join[{4}, LinearRecurrence[{2, 2, -1}, {5, 1, 16}, 30]] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    Vec((4-3*x-17*x^2+8*x^3)/(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) for n>3.
G.f.: (4 - 3 x - 17 x^2 + 8 x^3)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(-9*(-1)^n*2^(1+n) + (3+sqrt(5))^n*(-1+2*sqrt(5)) - (3-sqrt(5))^n*(1+2*sqrt(5))))/5 for n>0. - Colin Barker, Oct 20 2016

A266700 Coefficient of x in minimal polynomial of the continued fraction [1^n,1/2,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

0, -10, -12, -44, -102, -280, -720, -1898, -4956, -12988, -33990, -89000, -232992, -609994, -1596972, -4180940, -10945830, -28656568, -75023856, -196415018, -514221180, -1346248540, -3524524422, -9227324744, -24157449792, -63245024650, -165577624140
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:
[1/2,1,1,1,1,...] = sqrt(5)/2 has p(0,x) = -5 + 4*x^2, so a(0) = 0;
[1,1/2,1,1,1,...] = (5 + 2*sqrt(5))/5 has p(1,x) = 1 - 10*x + 5*x^2, so a(1) = -10;
[1,1,1/2,1,1,...] = 6 - 2*sqrt(5) has p(2,x) = 16 - 12*x + x^2, so a(2) = -12.

Links

Crossrefs

Cf. A265762, A266699.

Programs

  • Magma
    I:=[0,-10,-12]; [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], {1/2}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266699 *)
Coefficient[t, x, 1] (* A266700 *)
Coefficient[t, x, 2] (* A266699 *)
LinearRecurrence[{2, 2, -1}, {0, -10, -12}, 30] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    concat(0, Vec((-10*x+8*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.: (2 (-5 x + 4 x^2))/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(9*(-1)^n*2^(1+n) + (3-sqrt(5))^n*(-9+sqrt(5)) - (3+sqrt(5))^n*(9+sqrt(5))))/5. - Colin Barker, Oct 20 2016

A266701 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,1/3,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

9, 11, 5, 41, 81, 239, 599, 1595, 4149, 10889, 28481, 74591, 195255, 511211, 1338341, 3503849, 9173169, 24015695, 62873879, 164605979, 430944021, 1128226121, 2953734305, 7732976831, 20245196151, 53002611659, 138762638789, 363285304745, 951093275409
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:
[1/3,1,1,1,...] = (-1 + 3 sqrt(5))/6 has p(0,x) = -11 + 3 x + 9 x^2, so a(0) = 9;
[1,1/3,1,1,...] = (25 + 9 sqrt(5))/22 has p(1,x) = 5 - 25 x + 11 x^2, so a(1) = 11;
[1,1,1/3,1,...] = (35 - 9 sqrt(5))/10 has p(2,x) = 41 - 35 x + 5 x^2, so a(2) = 5.

Links

Crossrefs

Cf. A265762, A266702.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/3}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266701 *)
Coefficient[t, x, 1] (* A266702 *)
Coefficient[t, x, 2] (* A266701 *)
  • PARI
    a(n) = round((2^(-n)*(-37*(-2)^n-2*(3-sqrt(5))^n*(2+3*sqrt(5))+(3+sqrt(5))^n*(-4+6*sqrt(5))))/5) \\ Colin Barker, Sep 29 2016
  • PARI
    Vec((9-7*x-35*x^2+18*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: (9 - 7 x - 35 x^2 + 18 x^3)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(-37*(-2)^n-2*(3-sqrt(5))^n*(2+3*sqrt(5))+(3+sqrt(5))^n*(-4+6*sqrt(5))))/5. - Colin Barker, Sep 29 2016

Extensions

Three typos in data fixed by Colin Barker, Sep 29 2016

A266702 Coefficient of x in minimal polynomial of the continued fraction [1^n,1/3,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

3, -25, -35, -117, -279, -757, -1955, -5145, -13443, -35221, -92183, -241365, -631875, -1654297, -4330979, -11338677, -29685015, -77716405, -203464163, -532676121, -1394564163, -3651016405, -9558485015, -25024438677, -65514830979, -171520054297
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:
[1/3,1,1,1,...] = (-1 + 3 sqrt(5))/6 has p(0,x) = -11 + 3 x + 9 x^2, so a(0) = 9;
[1,1/3,1,1,...] = (25 + 9 sqrt(5))/22 has p(1,x) = 5 - 25 x + 11 x^2, so a(1) = 11;
[1,1,1/3,1,...] = (35 - 9 sqrt(5))/10 has p(2,x) = 41 - 35 x + 5 x^2, so a(2) = 5.

Links

Crossrefs

Cf. A265762, A266701.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {1/3}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266701 *)
Coefficient[t, x, 1] (* A266702 *)
Coefficient[t, x, 2] (* A266701 *)
  • PARI
    Vec((3-31*x+9*x^2+6*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: (3 - 31 x + 9 x^2 + 6 x^3)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(37*(-2)^n+2*(3-sqrt(5))^n*(-13+sqrt(5))-2*(3+sqrt(5))^n*(13+sqrt(5))))/5 for n>0. - Colin Barker, Sep 29 2016

A266703 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

9, 11, 1, 29, 45, 149, 359, 971, 2511, 6605, 17261, 45221, 118359, 309899, 811295, 2124029, 5560749, 14558261, 38113991, 99783755, 261237231, 683927981, 1790546669, 4687712069, 12272589495, 32130056459, 84117579839, 220222683101, 576550469421, 1509428725205
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:
[2/3,1,1,1,...] = (1+3*sqrt(5))/6 has p(0,x) = -11 - 3 x + 9 x^2, so a(0) = 9;
[1,2/3,1,1,...] = (19+9*sqrt(5))/22 has p(1,x) = -1 - 19 x + 11 x^2, so a(1) = 11;
[1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.

Links

Crossrefs

Cf. A265762, A266704.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266703 *)
Coefficient[t, x, 1] (* A266704 *)
Coefficient[t, x, 2] (* A266703 *)
  • PARI
    Vec((9-7*x-39*x^2+14*x^3-4*x^4+2*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: (9 - 7 x - 39 x^2 + 14 x^3 - 4 x^4 + 2 x^5)/(1 - 2 x - 2 x^2 + x^3).
a(n) = 2^(-n)*(-43*(-2)^n+(3+sqrt(5))^n*(-1+3*sqrt(5))-(3-sqrt(5))^n*(1+3*sqrt(5)))/5 for n>2. - Colin Barker, Sep 29 2016

A266704 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,2/3,1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

-3, -19, 17, -75, -165, -463, -1181, -3123, -8145, -21355, -55877, -146319, -383037, -1002835, -2625425, -6873483, -17994981, -47111503, -123339485, -322906995, -845381457, -2213237419, -5794330757, -15169754895, -39714933885, -103975046803, -272210206481
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:
[2/3,1,1,1,...] = (1+3*sqrt(5))/6 has p(0,x) = -11 - 3 x + 9 x^2, so a(0) = 9;
[1,2/3,1,1,...] = (19+9*sqrt(5))/22 has p(1,x) = -1 - 19 x + 11 x^2, so a(1) = 11;
[1,1,2/3,1,...] = (-17+9*sqrt(5))/2 has p(2,x) = -29 + 17 x + x^2, so a(2) = 1.

Links

Crossrefs

Cf. A265762, A266703.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2/3}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266703 *)
Coefficient[t, x, 1] (* A266704 *)
Coefficient[t, x, 2] (* A266703 *)
  • PARI
    Vec(-(3+13*x-61*x^2+74*x^3+68*x^4-34*x^5)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: (-3 - 13 x + 61 x^2 - 74 x^3 - 68 x^4 + 34 x^5)/(1 - 2 x - 2 x^2 + x^3).
a(n) = 2^(-n)*(43*(-2)^n+2*(3-sqrt(5))^n*(-7+sqrt(5))-2*(3+sqrt(5))^n*(7+sqrt(5)))/5 for n>2. - Colin Barker, Sep 29 2016

A266705 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,sqrt(5),1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

1, 11, 11, 45, 101, 281, 719, 1899, 4955, 12989, 33989, 89001, 232991, 609995, 1596971, 4180941, 10945829, 28656569, 75023855, 196415019, 514221179, 1346248541, 3524524421, 9227324745, 24157449791, 63245024651, 165577624139, 433487847789, 1134885919205
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(5),1,1,1,1,...] = (-1+3*sqrt(5))/2 has p(0,x)=-11+x+x^2, so a(0) = 1;
[1,sqrt(5),1,1,1,...] = (23+3*sqrt(5))/22 has p(1,x)=11-23x+11x^2, so a(1) = 11;
[1,1,sqrt(5),1,1,...] = (45-3*sqrt(5))/22 has p(2,x)=45-45x+11x^2, so a(2) = 11.

Links

Crossrefs

Cf. A265762, A266706.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[5]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266705 *)
Coefficient[t, x, 1] (* A266706 *)
Coefficient[t, x, 2] (* A266705 *)
  • PARI
    Vec((1+9*x-13*x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3.
G.f.: (1 + 9*x - 13*x^2 + 2*x^3)/(1 - 2*x - 2*x^2 + x^3).
a(n) = (2^(-n)*(-23*(-2)^n-(3-sqrt(5))^n*(-9+sqrt(5))+(3+sqrt(5))^n*(9+sqrt(5))))/5 for n>0. - Colin Barker, Sep 29 2016

A266706 Coefficient of x in minimal polynomial of the continued fraction [1^n,sqrt(5),1,1,1,...], where 1^n means n ones.

Original entry on oeis.org

1, -23, -45, -135, -337, -899, -2337, -6135, -16045, -42023, -110001, -288003, -753985, -1973975, -5167917, -13529799, -35421457, -92734595, -242782305, -635612343, -1664054701, -4356551783, -11405600625, -29860250115, -78175149697, -204665198999
Offset: 1

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(5),1,1,1,1,...] = (-1+3*sqrt(5))/2 has p(0,x)=-11+x+x^2, so a(0) = 1;
[1,sqrt(5),1,1,1,...] = (23+3*sqrt(5))/22 has p(1,x)=11-23x+11x^2, so a(1) = 11;
[1,1,sqrt(5),1,1,...] = (45-3* sqrt(5))/22 has p(2,x)=45-45x+11x^2, so a(2) = 11.

Links

Crossrefs

Cf. A265762, A266705.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[5]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266705 *)
Coefficient[t, x, 1] (* A266706 *)
Coefficient[t, x, 2] (* A266705 *)
LinearRecurrence[{2,2,-1},{1,-23,-45,-135},40] (* Harvey P. Dale, Jul 30 2017 *)
  • PARI
    Vec(x*(1-25*x-x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: x*(1 - 25*x - x^2 + 2*x^3) / ((1 + x)*(1 - 3*x + x^2)).
a(n) = (1/5)*2^(-n)*(-23*(-2)^n + (4-8*sqrt(5)) * (3+sqrt(5))^n + (3-sqrt(5))^n*(4+8*sqrt(5))) for n>1. - Colin Barker, May 21 2020
a(n) = 3*a(n-1) - a(n-2) - 23*(-1)^n for n >= 4. - Greg Dresden, May 18 2020

A266707 Coefficient of x^2 in minimal polynomial of the continued fraction [1^n,tau,1,1,1,...], where 1^n means n ones and tau = golden ratio = (1 + sqrt(5))/2.

Original entry on oeis.org

1, 5, 4, 19, 41, 116, 295, 781, 2036, 5339, 13969, 36580, 95759, 250709, 656356, 1718371, 4498745, 11777876, 30834871, 80726749, 211345364, 553309355, 1448582689, 3792438724, 9928733471, 25993761701, 68052551620, 178163893171, 466439127881, 1221153490484
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:
[tau,1,1,1,1,...] = sqrt(5) has p(0,x) = -5 + x^2, so a(0) = 1;
[1,tau,1,1,1,...] = (5 + sqrt(5))/5 has p(1,x) = 4 - 10 x + 5 x^2, so a(1) = 5;
[1,1,tau,1,1,...] = (9 - sqrt(5))/4 has p(2,x) = 19 - 18 x + 4 x^2, so a(2) = 4.

Links

Crossrefs

Cf. A265762, A266708.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {GoldenRatio}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A266707 *)
Coefficient[t, x, 1] (* A266708 *)
Coefficient[t, x, 2] (* A266707 *)
  • PARI
    Vec((1+3*x-8*x^2+2*x^3)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

Formula

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.: (1 + 3 x - 8 x^2 + 2 x^3)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(3+n)-(3-sqrt(5))^n*(-7+sqrt(5))+(3+sqrt(5))^n*(7+sqrt(5))))/5 for n>0. - Colin Barker, Sep 29 2016
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