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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A267079 Coefficient of x in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-15, 45, -633, -12321, -212379, -3867255, -68998575, -1240820397, -22247101689, -399334774401, -7164902653275, -128574917201655, -2307142450214223, -41400271270803501, -742895806968482169, -13330737506206610145, -239210288473732159515
Offset: 0

Views

Author

Clark Kimberling, Jan 11 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x)  = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = -15.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 45;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = -633.

Links

Crossrefs

Cf. A265762, A267078, A267080, A267081, A267082, A267083, A266527.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
  • PARI
    Vec(-3*(5 - 80*x - 114*x^2 + 4224*x^3 - 7142*x^4 - 7912*x^5 + 3123*x^6 + 390*x^7 - 30*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30)) \\ Andrew Howroyd, Mar 07 2018

Formula

a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (3 (-5 + 80 x + 114 x^2 - 4224 x^3 + 7142 x^4 + 7912 x^5 - 3123 x^6 - 390 x^7 + 30 x^8))/(1 - 13 x - 104 x^2 + 260 x^3 + 260 x^4 - 104 x^5 - 13 x^6 + x^7).
G.f.: -3*(5 - 80*x - 114*x^2 + 4224*x^3 - 7142*x^4 - 7912*x^5 + 3123*x^6 + 390*x^7 - 30*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)). - Andrew Howroyd, Mar 07 2018

A267080 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-6, -66, 1110, 18318, 333750, 5938446, 106865274, 1915425570, 34385669382, 616923941070, 11070947149014, 198655308975486, 3564757609030650, 63966755470710018, 1147838391054195510, 20597113658105850126, 369600280281802257654, 6632207432249371045230
Offset: 0

Views

Author

Clark Kimberling, Jan 11 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x)  = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = -6.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = -66;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = 1110.

Links

Crossrefs

Cf. A265762, A267078, A267079, A267081, A267082, A267083, A266527.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
  • PARI
    Vec(-6*(1 - 3*x - 429*x^2 - 1103*x^3 + 7527*x^4 - 1975*x^5 - 308*x^6 + 22*x^7)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30)) \\ Andrew Howroyd, Mar 07 2018

Formula

a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: -((6 (1 - 3 x - 429 x^2 - 1103 x^3 + 7527 x^4 - 1975 x^5 - 308 x^6 + 22 x^7))/(1 - 14 x - 90 x^2 + 350 x^3 - 90 x^4 - 14 x^5 + x^6)).
From Andrew Howroyd, Mar 07 2018: (Start)
a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6) for n > 7.
G.f.: -6*(1 - 3*x - 429*x^2 - 1103*x^3 + 7527*x^4 - 1975*x^5 - 308*x^6 + 22*x^7)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)).
(End)

A267081 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-9, 35, -969, -14359, -279261, -4862231, -88270665, -1576950691, -28345226121, -508305487319, -9123426587229, -163697793422935, -2937543639603849, -52711355807057699, -945871877489577801, -16972948054702729111, -304567428780675699165, -5465239154667149397911
Offset: 0

Views

Author

Clark Kimberling, Jan 11 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x)  = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = -9.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 35;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = -969.

Links

Crossrefs

Cf. A265762, A267078, A267079, A267080, A267082, A267083, A266527.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
  • PARI
    Vec((-9 + 125*x - 104*x^2 - 8179*x^3 - 9491*x^4 - 700*x^5 + 70*x^6)/((1 + x)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30)) \\ Andrew Howroyd, Mar 07 2018

Formula

a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (-9 + 125 x - 104 x^2 - 8179 x^3 - 9491 x^4 - 700 x^5 + 70 x^6)/(1 - 10 x - 135 x^2 - 135 x^3 - 10 x^4 + x^5).
From Andrew Howroyd, Mar 07 2018: (Start)
a(n) = 10*a(n-1) + 135*a(n-2) + 135*a(n-3) + 10*a(n-4) - a(n-5) for n > 6.
G.f.: (-9 + 125*x - 104*x^2 - 8179*x^3 - 9491*x^4 - 700*x^5 + 70*x^6)/((1 + x)*(1 + 7*x + x^2)*(1 - 18*x + x^2)).
(End)

A267082 Coefficient of x^4 in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

0, 6, 456, 6240, 131238, 2238780, 41011296, 730283034, 13143304440, 235581102912, 4229156006790, 75876624195564, 1361636473680576, 24432987781993530, 438436202143461288, 7867390833380267040, 141174789462751501926, 2533277512666920359964
Offset: 0

Views

Author

Clark Kimberling, Jan 11 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x)  = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = 0.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 6;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = 456.

Links

Crossrefs

Cf. A265762, A267078, A267079, A267080, A267081, A267083, A266527.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
  • PARI
    concat([0], 6*Vec((1 + 62*x - 114*x^2 + 823*x^3 - 182*x^4 - 28*x^5 + 2*x^6)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30))) \\ Andrew Howroyd, Mar 07 2018

Formula

a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (6 (x + 62 x^2 - 114 x^3 + 823 x^4 - 182 x^5 - 28 x^6 + 2 x^7))/(1 - 14 x - 90 x^2 + 350 x^3 - 90 x^4 - 14 x^5 + x^6).
From Andrew Howroyd, Mar 07 2018: (Start)
a(n) = 14*a(n-1) + 90*a(n-2) - 350*a(n-3) + 90*a(n-4) + 14*a(n-5) - a(n-6) for n > 7.
G.f.: 6*x*(1 + 62*x - 114*x^2 + 823*x^3 - 182*x^4 - 28*x^5 + 2*x^6)/((1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)).
(End)

A266527 Coefficient of x^6 in the minimal polynomial of the continued fraction [1^n,2^(1/3),1,1,...], where 1^n means n ones.

Original entry on oeis.org

1, 5, 11, 131, 3421, 56209, 1049105, 18561659, 334918459, 5997328339, 107703879581, 1932077585345, 34673771913121, 622167861459451, 11164539354582251, 200338227165577379, 3594932551574173405, 64508386001097153649, 1157556438367284595889
Offset: 0

Views

Author

Clark Kimberling, Jan 11 2016

Keywords

Comments

See A265762 for a guide to related sequences.

Examples

			Let u = 2^(1/3), and let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[u,1,1,1,...] has p(0,x)  = -5 - 15 x - 6 x^2 - 9 x^3 + 3 x^5 + x^6, so that a(0) = 1.
[1,u,1,1,1,...] has p(1,x) = -11 + 45 x - 66 x^2 + 35 x^3 + 6 x^4 - 15 x^5 + 5 x^6, so that a(1) = 5;
[1,1,u,1,1,1...] has p(2,x) = 131 - 633 x + 1110 x^2 - 969 x^3 + 456 x^4 - 111 x^5 + 11 x^6, so that a(2) = 11.

Links

Crossrefs

Cf. A265762, A267078, A267079, A267080, A267081, A267082, A267083.

Programs

  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2^(1/3)}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 30}]
Coefficient[t, x, 0]; (* A267078 *)
Coefficient[t, x, 1]; (* A267079 *)
Coefficient[t, x, 2]; (* A267080 *)
Coefficient[t, x, 3]; (* A267081 *)
Coefficient[t, x, 4]; (* A267082 *)
Coefficient[t, x, 5]; (* A267083 *)
Coefficient[t, x, 6]; (* A266527 *)
  • PARI
    Vec((1 - 8*x - 158*x^2 - 272*x^3 + 2134*x^4 + 2168*x^5 - 1009*x^6 - 130*x^7 + 10*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)) + O(x^30)) \\ Andrew Howroyd, Mar 07 2018

Formula

a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7) for n > 8.
G.f.: (1 - 8 x - 158 x^2 - 272 x^3 + 2134 x^4 + 2168 x^5 - 1009 x^6 - 130 x^7 + 10 x^8)/(1 - 13 x - 104 x^2 + 260 x^3 + 260 x^4 - 104 x^5 - 13 x^6 + x^7).
G.f.: (1 - 8*x - 158*x^2 - 272*x^3 + 2134*x^4 + 2168*x^5 - 1009*x^6 - 130*x^7 + 10*x^8)/((1 + x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2)). - Andrew Howroyd, Mar 07 2018

A266710 Coefficient of x^0 in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-1, 1, -9, 31, 311, 1889, 13599, 91519, 631721, 4318271, 29628279, 202995649, 1391561279, 9537357311, 65371447881, 448058829919, 3071050697399, 21049268992289, 144273903091551, 988867867179391, 6777801652728809, 46455742430697599, 318412398690263799
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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = -1;
[1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = 1;
[1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = -9.

Links

Crossrefs

Cf. A265762, A266711, A266712, A266713.

Programs

  • Magma
    I:=[31,311,1889,13599,91519]; [-1,1,-9] cat [n le 5 select I[n] else 5*Self(n-1) +15*Self(n-2) -15*Self(n-3) -5*Self(n-4) + Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 26 2018
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266710 *)
Coefficient[t, x, 1];  (* A266711 *)
Coefficient[t, x, 2];  (* A266712 *)
Coefficient[t, x, 3];  (* A266713 *)
Coefficient[t, x, 4];  (* A266710 *)
LinearRecurrence[{5,15,-15,-5,1},{-1,1,-9,31,311,1889,13599,91519},30] (* Harvey P. Dale, Jun 17 2016 *)
  • PARI
    x='x+O('x^30); Vec((1 -6*x -x^2 -46*x^3 -301*x^4 +260*x^5 +92*x^6 -18*x^7)/(-1 +5*x + 15*x^2 -15*x^3 -5*x^4 +x^5)) \\ G. C. Greubel, Jan 26 2018

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (1 -6*x -x^2 -46*x^3 -301*x^4 +260*x^5 +92*x^6 -18*x^7)/(-1 +5*x + 15*x^2 -15*x^3 -5*x^4 +x^5).

A266711 Coefficient of x in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-6, 2, 18, -102, -714, -4826, -33222, -227298, -1558962, -10682534, -73226346, -501882042, -3439999878, -23577981122, -161606223954, -1107664654566, -7592048797962, -52036670543258, -356664661728582, -2444615917773474, -16755646877311986, -114844911923314982
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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = -6;
[1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = 2;
[1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = 18.

Links

Crossrefs

Cf. A265762, A266710, A266712, A266713.

Programs

  • Magma
    I:=[-102, -714, -4826, -33222, -227298]; [-6, 2, 18] cat [n le 5 select I[n] else 5*Self(n-1) + 15*Self(n-2) - 15*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 26 2018
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266710 *)
Coefficient[t, x, 1];  (* A266711 *)
Coefficient[t, x, 2];  (* A266712 *)
Coefficient[t, x, 3];  (* A266713 *)
Coefficient[t, x, 4];  (* A266710 *)
LinearRecurrence[{5,15,-15,-5,1}, {-6, 2, 18, -102, -714, -4826, -33222, -227298}, 30] (* G. C. Greubel, Jan 26 2018 *)
  • PARI
    x='x+O('x^30); Vec(2*(3 -16*x -49*x^2 +156*x^3 +237*x^4 -280*x^5 -88*x^6 +18*x^7)/(-1 +5*x +15*x^2 -15*x^3 -5*x^4 +x^5)) \\ G. C. Greubel, Jan 26 2018

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*(3 -16*x -49*x^2 +156*x^3 +237*x^4 -280*x^5 -88*x^6 +18*x^7)/(-1 +5*x +15*x^2 -15*x^3 -5*x^4 +x^5).

A266712 Coefficient of x^2 in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.

Original entry on oeis.org

-5, -7, -7, 115, 607, 4615, 30427, 211687, 1442695, 9909907, 67867135, 465315847, 3188935867, 21858303175, 149816390407, 1026863749555, 7038210692767, 48240661271047, 330646286854555, 2266283690589607, 15533338646986375, 106467089195295187
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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = -5;
[1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = -7;
[1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = -7.

Links

Crossrefs

Cf. A265762, A266710, A266711, A266713.

Programs

  • Magma
    I:=[115, 607, 4615, 30427, 211687]; [-5, -7, -7] cat [n le 5 select I[n] else 5*Self(n-1) + 15*Self(n-2) - 15*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 26 2018
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266710 *)
Coefficient[t, x, 1];  (* A266711 *)
Coefficient[t, x, 2];  (* A266712 *)
Coefficient[t, x, 3];  (* A266713 *)
Coefficient[t, x, 4];  (* A266710 *)
LinearRecurrence[{5,15,-15,-5,1}, {-5, -7, -7, 115, 607, 4615, 30427, 211687}, 30] (* G. C. Greubel, Jan 26 2018 *)
  • PARI
    x='x+O('x^30); Vec((5 -18*x -103*x^2 -180*x^3 -7*x^4 +280*x^5 +56*x^6 -14*x^7)/(-1 + 5*x +15*x^2 -15*x^3 -5*x^4 +x^5)) \\ G. C. Greubel, Jan 26 2018

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: (5 -18*x -103*x^2 -180*x^3 -7*x^4 +280*x^5 +56*x^6 -14*x^7)/(-1 + 5*x +15*x^2 -15*x^3 -5*x^4 +x^5).

A266713 Coefficient of x^3 in the minimal polynomial of the continued fraction [1^n,sqrt(2),1,1,...], where 1^n means n ones.

Original entry on oeis.org

2, 2, -2, -54, -226, -1958, -12382, -87618, -593374, -4085846, -27955618, -191739462, -1313864638, -9006244994, -61727410366, -423092015478, -2899899974242, -19876251587558, -136233746512414, -933760274094786, -6400087386491038, -43866853488227222
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(2),1,1,1,...] has p(0,x) = -1 - 6 x - 5 x^2 + 2 x^3 + x^4, so a(0) = 2;
[1,sqrt(2),1,1,1,...] has p(1,x) = 1 + 2 x - 7 x^2 + 2 x^3 + x^4, so a(1) = 2;
[1,1,sqrt(2),1,1,1...] has p(2,x) = -9 + 18 x - 7 x^2 - 2 x^3 + x^4, so a(2) = -2.

Links

Crossrefs

Cf. A265762, A266710, A266711, A266712.

Programs

  • Magma
    I:=[-54, -226, -1958, -12382, -87618]; [2, 2, -2] cat [n le 5 select I[n] else 5*Self(n-1) + 15*Self(n-2) - 15*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 26 2018
  • Mathematica
    u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {Sqrt[2]}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 40}];
Coefficient[t, x, 0] ; (* A266710 *)
Coefficient[t, x, 1];  (* A266711 *)
Coefficient[t, x, 2];  (* A266712 *)
Coefficient[t, x, 3];  (* A266713 *)
Coefficient[t, x, 4];  (* A266710 *)
LinearRecurrence[{5,15,-15,-5,1}, {2, 2, -2, -54, -226, -1958, -12382, -87618}, 30] (* G. C. Greubel, Jan 26 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec(-2*(1 -4*x -21*x^2 -22*x^3 +57*x^4 -20*x^5 -12*x^6 +2*x^7)/(-1 +5*x +15*x^2 -15*x^3 -5*x^4 +x^5)) \\ G. C. Greubel, Jan 26 2018

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 -4*x -21*x^2 -22*x^3 +57*x^4 -20*x^5 -12*x^6 +2*x^7)/(-1 +5*x +15*x^2 -15*x^3 -5*x^4 +x^5).

A266799 Coefficient of x^0 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

1, 1, 49, 229, 1861, 12001, 84241, 572209, 3935569, 26939221, 184737301, 1265964481, 8677687969, 59476087009, 407659540081, 2794128600901, 19151272325221, 131264694791329, 899701808208049, 6166647394567441, 42266831441062801, 289701168799073461
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) = 1;
[1,sqrt(3),1,1,1,...] has p(1,x)=1-12x+23x^2-12x^3+x^4, so a(1) = 1;
[1,1,sqrt(3),1,1,1...] has p(2,x)=49-98x+65x^2-16x^3+x^4, so a(2) = 49.

Links

Crossrefs

Cf. A265762, A266800, 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 *)
LinearRecurrence[{5,15,-15,-5,1},{1,1,49,229,1861},30] (* Harvey P. Dale, Oct 21 2019 *)

Formula

a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5) .
G.f.: (-1 + 4 x - 29 x^2 + 16 x^3 - x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5).
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