OFFSET
0,2
COMMENTS
a(15+30k)-1 and a(15+30k)+1 are consecutive odd powerful numbers. The first pair is 13837575261124 +- 1. See A076445. - T. D. Noe, May 04 2006
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 15*y^2 = 1. The corresponding y values are in A001090. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 04 2014]
The square root of 15*(n^2-1) at those numbers = 5 * A136325. - Richard R. Forberg, Nov 22 2013
For the above Diophantine equation x^2-15*y^2=1, x + y = A105426. - Richard R. Forberg, Nov 22 2013
a(n) solves for x in the Diophantine equation floor(3*x^2/5)= y^2. The corresponding y solutions are provided by A136325(n). x + y = A070997(n). - Richard R. Forberg, Nov 22 2013
Except for the first term, values of x (or y) in the solutions to x^2 - 8xy + y^2 + 15 = 0. - Colin Barker, Feb 05 2014
REFERENCES
Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
T. D. Noe, Table of n, a(n) for n=0..200
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (8,-1).
FORMULA
G.f.: A(x) = (1-4*x)/(1-8*x+x^2). - Simon Plouffe in his 1992 dissertation
For all elements x of the sequence, 15*(x^2 -1) is a square. Limit_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 11 2002
a(n) = sqrt(15*((A001090(n))^2)+1).
a(n) = ((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 4*S(n-1, 8) - S(n-2, 8) = (S(n, 8) - S(n-2, 8))/2, n>=1; S(n, x) := U(n, x/2) with Chebyshev's polynomials of the 2nd kind, A049310, with S(-1, x) := 0 and S(-2, x) := -1.
a(n) = T(n, 4) with Chebyshev's polynomials of the first kind; see A053120.
a(n)=a(-n). - Ralf Stephan, Jun 06 2005
a(n)*a(n+3) - a(n+1)*a(n+2) = 120. - Ralf Stephan, Jun 06 2005
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*15^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 15*x^2))^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 16*x + 4*x^2) is the g.f. of A098269.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - (5/2)/a(n)) = 1/3.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (3/2)/a(n)) = 1/5.
Sum_{n >= 1} 1/(a(n)^2 - 5/2) = 1/3 - 1/sqrt(15). (End)
MATHEMATICA
LinearRecurrence[{8, -1}, {1, 4}, 20] (* Harvey P. Dale, May 01 2014 *)
PROG
(PARI) a(n)=subst(poltchebi(n), x, 4)
(PARI) a(n)=n=abs(n); polcoeff((1-4*x)/(1-8*x+x^2)+x*O(x^n), n) /* Michael Somos, Jun 07 2005 */
(Magma) R<x>:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1-4*x)/(1-8*x+x^2) )); // G. C. Greubel, Aug 26 2019
(Sage)
def A001091_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P( (1-4*x)/(1-8*x+x^2) ).list()
A001091_list(20) # G. C. Greubel, Aug 26 2019
(GAP) a:=[1, 4];; for n in [3..20] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 26 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Aug 25 2000
Chebyshev comments from Wolfdieter Lang, Oct 31 2002
STATUS
approved