OFFSET
0,2
COMMENTS
Chebyshev's polynomials U(n,x) evaluated at x=6.
a(n) give all (nontrivial, integer) solutions of Pell equation b(n)^2 - 35*a(n)^2 = +1 with b(n)=A023038(n+1), n>=0.
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 12's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,11}. - Milan Janjic, Jan 26 2015
a(n) = -a(-2-n) for all n in Z. - Michael Somos, Jun 29 2019
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..900
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (12,-1).
FORMULA
a(n) = S(n, 12) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310.
a(n) = ((6+sqrt(35))^(n+1) - (6-sqrt(35))^(n+1))/(2*sqrt(35)).
a(n) = sqrt((A023038(n)^2 - 1)/35).
[A077417(n), a(n)] = the 2 X 2 matrix [1,10; 1,11]^(n+1) * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n) = 12*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=12. - Philippe Deléham, Nov 17 2008
a(n) = b such that (-1)^(n+1)*Integral_{x=0..Pi/2} (sin((n+1)*x))/(6+cos(x)) dx = c + b*(log(2)+log(3)-log(7)). - Francesco Daddi, Aug 01 2011
a(n) = Sum_{k=0..n} A101950(n,k)*11^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012 (Start):
Product_{n>=0} (1 + 1/a(n)) = 1/5*(5 + sqrt(35)).
Product_{n>=1} (1 - 1/a(n)) = 1/12*(5 + sqrt(35)). (End)
E.g.f.: exp(6*x)*(35*cosh(sqrt(35)*x) + 6*sqrt(35)*sinh(sqrt(35)*x))/35. - Stefano Spezia, Dec 14 2022
EXAMPLE
G.f. = 1 + 12*x + 143*x^2 + 1704*x^3 + 20305*x^4 + 241956*x^5 + ...
MAPLE
seq( simplify(ChebyshevU(n, 6)), n=0..20); # G. C. Greubel, Dec 23 2019
MATHEMATICA
Table[GegenbauerC[n, 1, 6], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[1/(1-12*x+x^2), {x, 0, 30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{12, -1}, {1, 12}, 30] (* Harvey P. Dale, Feb 17 2016 *)
a[n_]:= ChebyshevU[n, 6]; (* Michael Somos, Jun 29 2019 *)
PROG
(Sage) [lucas_number1(n, 12, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
(Sage) [chebyshev_U(n, 6) for n in (0..20)] # G. C. Greubel, Dec 23 2019
(Magma) I:=[1, 12]; [n le 2 select I[n] else 12*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jun 13 2012
(PARI) Vec(1/(1-12*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012
(PARI) {a(n) = polchebyshev(n, 2, 6)}; \\ Michael Somos, Jun 29 2019
(GAP) m:=8;; a:=[1, 2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
CROSSREFS
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), this sequence (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Chebyshev comments and a(n) formulas from Wolfdieter Lang, Nov 08 2002
STATUS
approved