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 1-5 of 5 results.

A087798 a(n) = 9*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 9.

Original entry on oeis.org

2, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, 432083484, 3936182123, 35857722591, 326655685442, 2975758891569, 27108485709563, 246952130277636, 2249677658208287, 20494051054152219, 186696137145578258
Offset: 0

Views

Author

Nikolay V. Kosinov, Dmitry V. Poljakov (kosinov(AT)unitron.com.ua), Oct 10 2003

Keywords

Comments

a(n+1)/a(n) converges to (9 + sqrt(85))/2.
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010

Examples

			a(4) = 9*a(3) + a(2) = 9*756 + 83 = 6887.

Links

Crossrefs

Cf. A014511.

Programs

  • Magma
    I:=[2,9]; [n le 2 select I[n] else 9*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016
  • Mathematica
    RecurrenceTable[{a[0] == 2, a[1] == 9, a[n] == 9 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{9,1}, {2,9}, 30] (* G. C. Greubel, Nov 07 2018 *)
  • PARI
    x='x+O('x^30); Vec((2-9*x)/(1-9*x-x^2)) \\ G. C. Greubel, Nov 07 2018

Formula

a(n) = ((9 + sqrt(85))/2)^n + ((9 - sqrt(85))/2)^n.
G.f.: (2 - 9*x)/(1 - 9*x - x^2). - Philippe Deléham, Nov 02 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 9*A097840(n), a(2n) = A099373(n).
a(3n+1) = A041150(5n), a(3n+2) = A041150(5n+3), a(3n+3) = 2*A041150(5n+4).
Lim_{k->infinity} a(n+k)/a(k) = (A087798(n) + A099371(n)*sqrt(85))/2.
Lim_{n->infinity} A087798(n)/A099371(n) = sqrt(85). (End)

Extensions

More terms from Ray Chandler, Nov 06 2003

A097837 Chebyshev polynomials S(n,51) + S(n-1,51) with Diophantine property.

Original entry on oeis.org

1, 52, 2651, 135149, 6889948, 351252199, 17906972201, 912904330052, 46540213860451, 2372638002552949, 120957997916339948, 6166485255730784399, 314369790044353664401, 16026692807006306100052, 817046963367277257438251
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(7*a(n))^2 - 53*b(n)^2 = -4 with b(n)=A097838(n) gives all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...

Links

Crossrefs

Cf. A097783, A097834, A097836, A097840.

Programs

  • GAP
    a:=[1,52];; for n in [3..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019
  • Mathematica
    LinearRecurrence[{51,-1}, {1,52}, 30] (* G. C. Greubel, Jan 12 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1+x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019
  • Sage
    ((1+x)/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019

Formula

a(n) = S(n, 51) + S(n-1, 51) = S(2*n, sqrt(53)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 51)=A097836(n).
a(n) = (-2/7)*i*((-1)^n)*T(2*n+1, 7*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-51*x+x^2).
a(n) = 51*a(n-1) - a(n-2); a(0)=1, a(1)=52. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/7)*(7/2 o 7/2 o ... o 7/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 52, 0, 2651, 0, 135149, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 51*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -49, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,7) + (1/7)*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 14*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*x^n.
Exp( Sum_{n >= 1} (-14)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*(-x)^n. (End)

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Original entry on oeis.org

1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) = (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)

Examples

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...

Links

Crossrefs

A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008
Cf. A097781, A097783, A097837, A097840.

Programs

  • Mathematica
    a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)
  • PARI
    {a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

Formula

a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024

A097839 Chebyshev polynomials S(n,83).

Original entry on oeis.org

1, 83, 6888, 571621, 47437655, 3936753744, 326703123097, 27112422463307, 2250004361331384, 186723249568041565, 15495779709786118511, 1285962992662679794848, 106719432611292636853873, 8856426943744626179076611, 734976716898192680226504840
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

Used for all positive integer solutions of Pell equation x^2 - 85*y^2 = -4. See A097840 with A097841.

Links

Programs

  • GAP
    a:=[1,83];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019
  • Mathematica
    CoefficientList[Series[1/(1-83x+x^2),{x,0,20}],x] (* or *) LinearRecurrence[{83,-1},{1,83},20] (* Harvey P. Dale, Oct 11 2012 *)
  • PARI
    my(x='x+O('x^20)); Vec(1/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019
  • Sage
    (1/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

Formula

a(n) = S(n, 83) = U(n, 83/2) = S(2*n+1, sqrt(85))/sqrt(85) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x).
a(n) = 83*a(n-1) - a(n-2), n >= 1, a(-1)=0, a(0)=1, a(1)=83.
a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap = (83+9*sqrt(85))/2 and am = (83-9*sqrt(85))/2 = 1/ap.
G.f.: 1/(1-83*x+x^2).

Extensions

More terms from Harvey P. Dale, Oct 11 2012

A097841 First differences of Chebyshev polynomials S(n,83) = A097839(n) with Diophantine property.

Original entry on oeis.org

1, 82, 6805, 564733, 46866034, 3889316089, 322766369353, 26785719340210, 2222891938868077, 184473245206710181, 15309056460218076946, 1270467212952893676337, 105433469618629957059025
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(9*b(n))^2 - 85*a(n)^2 = -4 with b(n)=A097840(n) give all positive solutions of this Pell equation.
For n > 0, a(n) is the hypotenuse of the Pythagorean triple (x(n), y(n), a(n)) that is primitive for n == 0, 2 (mod 3) where (x(n)) and (y(n)) are recurrences of the form (82,82,-1) with the initial values x(0) = 1, x(1) = 80, x(2) = 6643; y(0) = 0, y(1) = 18, y(2) = 1476. - Klaus Purath, Jul 19 2025

Examples

			All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9=9*1,1), (756=9*84,82), (62739=9*6971,6805), (5206581=9*578509,564733), ...

Links

Crossrefs

Cf. A097839, A097840.
Cf. A049310, A053120.
Cf. A099371.

Programs

  • GAP
    a:=[1,82];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019
  • Mathematica
    CoefficientList[Series[(1-x)/(1-83x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{83,-1}, {1,82}, 20] (* G. C. Greubel, Jan 13 2019 *)
  • PARI
    my(x='x+O('x^20)); Vec((1-x)/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019
  • Sage
    ((1-x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

Formula

a(n) = ((-1)^n)*S(2*n, 9*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1 - 83*x + x^2).
a(n) = S(n, 83) - S(n-1, 83) = T(2*n+1, sqrt(85)/2)/(sqrt(85)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
a(n) = 83*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=82. - Philippe Deléham, Nov 18 2008
From Klaus Purath, Jul 19 2025: (Start)
a(n) = A099371(2n+1) = A099371(n)^2 + A099371(n+1)^2.
a(n) = (t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 82*t(i+2) + 82*t(i+1) - t(i) or t(i+2) = 83*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)
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