OFFSET
0,1
COMMENTS
Positive values of x satisfying x^2 - 21*y^2 = 4; values of y are in A004254. - Wolfdieter Lang, Nov 29 2002
Except for the first term, positive values of x (or y) satisfying x^2 - 5xy + y^2 + 21 = 0. - Colin Barker, Feb 08 2014
REFERENCES
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
Noureddine Chair, Exact two-point resistance, and the simple random walk on the complete graph minus N edges, Ann. Phys. 327, No. 12, 3116-3129 (2012), P(7).
Tanya Khovanova, Recursive Sequences
Lisa Lokteva, Constructing Rational Homology 3-Spheres That Bound Rational Homology 4-Balls, arXiv:2208.14850 [math.GT], 2022.
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
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-1).
FORMULA
a(n) = 5*S(n-1, 5) - 2*S(n-2, 5) = S(n, 5) - S(n-2, 5) = 2*T(n, 5/2), with S(n, x)=U(n, x/2), S(-1, x)=0, S(-2, x)=-1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 5) = A004254(n), n>=0.
G.f.: (2-5*x)/(1-5*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) ~ (1/2*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002
a(n) = ap^n + am^n, with ap=(5+sqrt(21))/2 and am=(5-sqrt(21))/2.
a(n) = sqrt(4 + 21*A004254(n)^2).
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(5 - sqrt(21)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.19827 65373 95327 17782 ... = 2 + 1/(5 + 1/(23 + 1/(110 + ...))).
Also F(-alpha) = 0.79824 49142 28050 93561 ... has the continued fraction representation 1 - 1/(5 - 1/(23 - 1/(110 - ...))) and the simple continued fraction expansion 1/(1 + 1/((5-2) + 1/(1 + 1/((23-2) + 1/(1 + 1/((110-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((5^2-4) + 1/(1 + 1/((23^2-4) + 1/(1 + 1/((110^2-4) + 1/(1 + ...))))))).
(End)
EXAMPLE
G.f. = 2 + 5*x + 23*x^2 + 110*x^3 + 527*x^4 + 2525*x^5 + ... - Michael Somos, Oct 25 2022
MAPLE
seq( simplify(2*ChebyshevT(n, 5/2)), n=0..30); # G. C. Greubel, Jan 16 2020
MATHEMATICA
a[0]=2; a[1]=5; a[n_]:= 5a[n-1] -a[n-2]; Table[a[n], {n, 0, 30}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{5, -1}, {2, 5}, 30] (* Harvey P. Dale, May 12 2019 *)
2*ChebyshevT[Range[0, 30], 5/2] (* G. C. Greubel, Jan 16 2020 *)
a[ n_] := LucasL[n, 5*I]/I^n; (* Michael Somos, Oct 25 2022 *)
PROG
(PARI) {a(n) = subst(poltchebi(n), x, 5/2)*2};
(PARI) {a(n) = polchebyshev(n, 1, 5/2)*2 }; /* Michael Somos, Oct 25 2022 */
(Sage) [lucas_number2(n, 5, 1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
(Magma) I:=[2, 5]; [n le 2 select I[n] else 5*Self(n-1) -Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2020
(Magma) R<x>:=PowerSeriesRing(Integers(), 25); Coefficients(R!((2-5*x)/(1-5*x+x^2))); // Marius A. Burtea, Jan 16 2020
(GAP) a:=[2, 5];; for n in [4..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 16 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Chebyshev comments from Wolfdieter Lang, Oct 31 2002
STATUS
approved