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A003501 a(n) = 5*a(n-1) - a(n-2).
(Formerly M1540)
12
2, 5, 23, 110, 527, 2525, 12098, 57965, 277727, 1330670, 6375623, 30547445, 146361602, 701260565, 3359941223, 16098445550, 77132286527, 369562987085, 1770682648898, 8483850257405, 40648568638127, 194758992933230 (list; graph; refs; listen; history; text; internal format)
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

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1

Tanya Khovanova, Recursive Sequences

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)

Index entries for sequences related to Chebyshev polynomials.

Index to sequences with 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)

a(n) = ( A217787(k+3n)+A217787(k-3n) )/A217787(k) for k>=3n. [Bruno Berselli, Mar 25 2013]

MATHEMATICA

a[0] = 2; a[1] = 5; a[n_] := 5a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 21}] (from Robert G. Wilson v Jan 30 2004)

PROG

(PARI) a(n)=if(n<0, 0, subst(poltchebi(n), x, 5/2)*2)

(Sage) [lucas_number2(n, 5, 1) for n in range(37)] - Zerinvary Lajos, Jun 25 2008

CROSSREFS

Cf. A004252, A004253, A217787.

Sequence in context: A219889 A100299 A038833 * A006990 A242227 A032182

Adjacent sequences:  A003498 A003499 A003500 * A003502 A003503 A003504

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane.

EXTENSIONS

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

STATUS

approved

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Last modified October 22 20:47 EDT 2014. Contains 248410 sequences.