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.

A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5*n(j)+1=a(j)*a(j) and 7*n(j)+1=b(j)*b(j) with positive integer numbers.

Original entry on oeis.org

0, 24, 3432, 487344, 69199440, 9825833160, 1395199109304, 198108447688032, 28130004372591264, 3994262512460271480, 567157146764985958920, 80532320578115545895184, 11435022364945642531157232, 1623692643501703123878431784, 230552920354876897948206156120
Offset: 1

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Author

Paul Weisenhorn, Apr 19 2009

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients( R!(24*x^2/((1-x)*(1-142*x+x^2)))); // G. C. Greubel, Jun 03 2018
    
  • Maple
    for a from 1 by 2 to 100000 do b:=sqrt((7*a*a-2)/5): if (trunc(b)=b) then
    n:=(a*a-1)/5: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
    # Second program
    seq((6/35)*(simplify(ChebyshevU(n,71) -141*ChebyshevU(n-1,71)) -1), n=1..30); # G. C. Greubel, Sep 27 2022
  • Mathematica
    LinearRecurrence[{143,-143,1}, {0, 24, 3432}, 30] (* or *) CoefficientList[Series[24*x^2/((1-x)*(1-142*x+x^2)), {x,0,30}], x] (* G. C. Greubel, Jun 03 2018 *)
  • PARI
    concat(0, Vec(-24*x^2/((x-1)*(x^2-142*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016
    
  • PARI
    a(n) = round((-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70) \\ Colin Barker, Jul 26 2016
    
  • SageMath
    [(6/35)*(-1 + chebyshev_U(n, 71) - 141*chebyshev_U(n-1, 71)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

Formula

The a(j) recurrence is a(1)=1, a(2)=11, a(t+2) = 12*a(t+1) - a(t) resulting in terms 1, 11, 131, 1561, ... (A077417).
The b(j) recurrence is b(1)=1, b(2)=13, b(t+2) = 12*b(t+1) - b(t) resulting in terms 1, 13, 155, 1847, ... (A077416).
The n(j) recurrence is n(0)=n(1)=0, n(2)=24, n(t+3) = 143*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 24, 3432, 487344, ... (this sequence).
G.f.: 24*x^2/((1-x)*(1-142*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70. - Colin Barker, Jul 26 2016
a(n) = (6/35)*(ChebyshevU(n, 71) - 141*ChebyshevU(n-1, 71) - 1). - G. C. Greubel, Sep 27 2022

Extensions

More terms from R. J. Mathar, Apr 20 2009