OFFSET
0,2
COMMENTS
Previous name was: The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 13*n(j) + 1 = a(j)*a(j) and 15*n(j) + 1 = b(j)*b(j) with positive integer numbers.
Positive values of x (or y) satisfying x^2 - 28*x*y + y^2 + 26 = 0. - Colin Barker, Feb 23 2014
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..199
Index entries for linear recurrences with constant coefficients, signature (28,-1).
FORMULA
G.f.: (1 - x)/(1 - 28*x + x^2).
The a(j) recurrence is a(0)=1, a(1)=27, a(t+2) = 28*a(t+1) - a(t) resulting in terms 1, 27, 755, 21113, ... (this sequence).
The b(j) recurrence is b(0)=1, b(1)=29, b(t+2) = 28*b(t+1) - b(t) resulting in terms 1, 29, 811, 22679, ... (A159669).
The n(j) recurrence is n(0) = n(1) = 0, n(2) = 56, n(t+3) = 783*(n(t+2) -n(t+1)) + n(t) resulting in terms 0, 0, 56, 43848, 34289136, ... (A159673).
a(n) = (1/30)*(15-sqrt(195))*(1+(14+sqrt(195))^(2*n+1))/(14+sqrt(195))^n. - Bruno Berselli, Feb 25 2014
a(n) = 28*a(n-1) - a(n-2), a(0)=1, a(1)=27. - Harvey P. Dale, Apr 09 2014
MAPLE
for a from 1 by 2 to 100000 do b:=sqrt((15*a*a-2)/13): if (trunc(b)=b) then
n:=(a*a-1)/13: La:=[op(La), a]:Lb:=[op(Lb), b]:Ln:=[op(Ln), n]: endif: enddo:
# Second program
seq(simplify(ChebyshevU(n, 14) -ChebyshevU(n-1, 14)), n=0..40); # G. C. Greubel, Sep 26 2022
MATHEMATICA
CoefficientList[Series[(1-x)/(1-28x+x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 25 2014 *)
LinearRecurrence[{28, -1}, {1, 27}, 40] (* Harvey P. Dale, Apr 09 2014 *)
PROG
(PARI) Vec((-x+1)/(x^2-28*x+1) + O(x^100)) \\ Colin Barker, Feb 23 2014
(Magma) [n le 2 select 27^(n-1) else 28*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 25 2014
(SageMath)
def A159668(n): return chebyshev_U(n, 14) - chebyshev_U(n-1, 14)
[A159668(n) for n in range(40)] # G. C. Greubel, Sep 26 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Apr 19 2009
EXTENSIONS
More terms from Colin Barker, Feb 23 2014
New name and offset changed to 0 from Joerg Arndt, Feb 23 2014
STATUS
approved