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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A159664 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11*n(j) + 1 = a(j)*a(j) and 13*n(j) + 1 = b(j)*b(j), with positive integer numbers.

Original entry on oeis.org

1, 23, 551, 13201, 316273, 7577351, 181540151, 4349386273, 104203730401, 2496540143351, 59812759710023, 1433009692897201, 34332419869822801, 822545067182850023, 19706749192518577751, 472139435553263016001, 11311639704085793806273, 271007213462505788334551
Offset: 1

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Author

Paul Weisenhorn, Apr 19 2009

Keywords

Comments

Positive values of x (or y) satisfying x^2 - 24*x*y + y^2 + 22 = 0. - Colin Barker, Feb 19 2014

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Magma
    [n le 2 select 23^(n-1) else 24*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 21 2014
    
  • Maple
    for a from 1 by 2 to 100000 do b:=sqrt((13*a*a-2)/11): if (trunc(b)=b) then
    n:=(a*a-1)/11: La:=[La),a]:Lb:=[op(Lb),b]: Ln:=[op(Ln),n]: end if: end do:
    # Second program
    seq(simplify(ChebyshevU(n-1,12) - ChebyshevU(n-2,12)), n=1..30); # G. C. Greubel, Sep 27 2022
  • Mathematica
    CoefficientList[Series[(1-x)/(1-24x+x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 21 2014 *)
    LinearRecurrence[{24,-1}, {1,23}, 30] (* G. C. Greubel, Sep 27 2022 *)
  • PARI
    Vec(x*(1-x)/(1-24*x+x^2) + O(x^100)) \\ Colin Barker, Feb 19 2014
    
  • PARI
    a(n) = round((12+sqrt(143))^(-n)*(13+sqrt(143)-(-13+sqrt(143))*(12+sqrt(143))^(2*n))/26) \\ Colin Barker, Jul 25 2016
    
  • SageMath
    def A159664(n): return chebyshev_U(n-1,12) - chebyshev_U(n-2,12)
    [A159664(n) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

Formula

The a(j) recurrence is a(1)=1, a(2)=23, a(t+2) = 24*a(t+1) - a(t) resulting in terms 1, 23, 551, 13201, ... (this sequence).
The b(j) recurrence is b(1)=1, b(2)=25, b(t+2) = 24*b(t+1) - b(t) resulting in terms 1, 25, 599, 14351, ... (A159661).
The n(j) recurrence is n(0)=n(1)=1, n(2)=48, n(t+3) = 575*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 48, 27600, 15842400, ... (A159665).
G.f.: x*(1-x)/(1 - 24*x + x^2). - Colin Barker, Feb 19 2014
a(n) = (12+sqrt(143))^(-n)*(13+sqrt(143)-(-13+sqrt(143))*(12+sqrt(143))^(2*n))/26. - Colin Barker, Jul 25 2016
a(n) = A077423(n-1) - A077423(n-2). - G. C. Greubel, Sep 27 2022

Extensions

More terms from Colin Barker, Feb 19 2014

A159665 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 11*n(j) + 1 = a(j)*a(j) and 13*n(j) + 1 = b(j)*b(j); with positive integer numbers.

Original entry on oeis.org

0, 48, 27600, 15842400, 9093510048, 5219658925200, 2996075129554800, 1719741904705530048, 987128857225844692800, 566610244305730148137200, 325233293102631879186060048, 186683343630666392922650330400, 107155914010709406905722103589600
Offset: 1

Views

Author

Paul Weisenhorn, Apr 19 2009

Keywords

Crossrefs

Programs

  • Magma
    I:=[0,48,27600]; [n le 3 select I[n] else 575*Self(n-1) -575*Self(n-2) +Self(n-3): n in [1..31]]; // G. C. Greubel, Jun 26 2022
    
  • Maple
    for a from 1 by 2 to 100000 do b:=sqrt((13*a*a-2)/11): if (trunc(b)=b) then
    n:=(a*a-1)/11: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: endif: enddo:
  • Mathematica
    LinearRecurrence[{575,-575,1}, {0,48,27600}, 30] (* G. C. Greubel, Jun 26 2022 *)
  • PARI
    concat(0, Vec(-48*x^2/((x-1)*(x^2-574*x+1)) + O(x^30))) \\ Colin Barker, Sep 25 2015
    
  • PARI
    a(n) = round((-24+(12+sqrt(143))*(287+24*sqrt(143))^(-n)-(-12+sqrt(143))*(287+24*sqrt(143))^n)/286) \\ Colin Barker, Jul 26 2016
    
  • SageMath
    [(12/143)*(chebyshev_U(n,287) -573*chebyshev_U(n-1,287) -1) for n in (1..30)] # G. C. Greubel, Jun 26 2022

Formula

The a(j) recurrence is a(1)=1; a(2)=23; a(t+2) = 24*a(t+1) - a(t) resulting in terms 1, 23, 551, 13201, ... (A159664).
The b(j) recurrence is b(1)=1; b(2)=25; b(t+2) = 24*b(t+1) - b(t) resulting in terms 1, 25, 599, 14351, ... (A159661).
The n(j) recurrence is n(0)=n(1)=1; n(2)=48; n(t+3) = 575*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 48, 27600, 15842400 as listed above.
From Colin Barker, Sep 25 2015: (Start)
a(n) = 575*a(n-1) - 575*a(n-2) + a(n-3) for n > 3.
G.f.: 48*x^2 / ((1-x)*(1-574*x+x^2)). (End)
a(n) = (-24 + (12 + sqrt(143))*(287 + 24*sqrt(143))^(-n) - (-12 + sqrt(143))*(287 + 24*sqrt(143))^n)/286. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 25 2022: (Start)
a(n) = (12/143)*(ChebyshevU(n, 287) - 573*ChebyshevU(n-1, 287) - 1).
E.g.f.: (12/143)*(exp(287*x)*( (sqrt(143)/12)*sinh(24*sqrt(143)*x) + cosh(24*sqrt(143)*x) ) - exp(x)). (End)
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