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.

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

Offset: 1

Author

Paul Weisenhorn, Apr 19 2009

Keywords

• nonn
• easy

Comments

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• Index entries for linear recurrences with constant coefficients, signature (24,-1).

Crossrefs

Cf. A077423, A157456, A159661, A159665.

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