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A003501
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a(n) = 5*a(n-1) - a(n-2).
(Formerly M1540)
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12
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2, 5, 23, 110, 527, 2525, 12098, 57965, 277727, 1330670, 6375623, 30547445, 146361602, 701260565, 3359941223, 16098445550, 77132286527, 369562987085, 1770682648898, 8483850257405, 40648568638127, 194758992933230
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OFFSET
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0,1
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COMMENTS
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Positive values of x satisfying x^2 - 21*y^2 = 4; values of y are in A004254. - Wolfdieter Lang, Nov 29 2002
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REFERENCES
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_Jeffrey Shallit_, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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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 S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for sequences related to linear recurrences with constant coefficients
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FORMULA
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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).
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]
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MAPLE
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A003501:=-(-2+5*z)/(1-5*z+z**2); [Conjectured by Simon Plouffe in his 1992 dissertation.]
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MATHEMATICA
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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)
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PROG
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(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 (zerinvarylajos(AT)yahoo.com), Jun 25 2008
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CROSSREFS
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Cf. A004252, A004253, A217787.
Sequence in context: A219889 A100299 A038833 * A006990 A032182 A165902
Adjacent sequences: A003498 A003499 A003500 * A003502 A003503 A003504
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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More terms from James A. Sellers, Aug 31 2000
Chebyshev comments from Wolfdieter Lang, Oct 31 2002
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STATUS
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approved
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