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A092499 Chebyshev polynomials S(n-1,21) with Diophantine property. 3
0, 1, 21, 440, 9219, 193159, 4047120, 84796361, 1776676461, 37225409320, 779956919259, 16341869895119, 342399310878240, 7174043658547921, 150312517518628101, 3149388824232642200, 65986852791366858099 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C.

The natural numbers a(n)=n satisfy the recurrence a(n-1)*a(n+1)=a(n)^2-1. Let R_r denote the sequence starting with 0,1,r and with this recurrence. We see that R_2 = "the natural numbers" and we find R_3 = A001906. These R_r form a "family" of sequences, which coincides with the m-family (r=m-2, n -> n+1) provided by Wolfdieter Lang (see A078368). This sequence R_21 is strongly related to A041833, which gives the denominators in the continued fraction of sqrt(437).

All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0.

For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 21'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,...,20}. Milan Janjic, Jan 25 2015

LINKS

Indranil Ghosh, Table of n, a(n) for n = 0..756

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (21,-1).

FORMULA

a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1)=a(n)^2-1

a(n)=S(n-1, 21)=U(n-1, 21/2) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x).

a(n)=S(2*n-1, sqrt(23))/sqrt(23), n>=1.

a(n)=21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1.

a(n)=(ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2.

G.f.: x/(1-21*x+x^2).

a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - Philippe Deléham, Feb 10 2012

Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - Peter Bala, Dec 23 2012

Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - Peter Bala, Dec 23 2012

EXAMPLE

a(3)=440 because a(1)*440=a(2)^2-1

MATHEMATICA

LinearRecurrence[{21, -1}, {0, 1}, 30] (* Harvey P. Dale, Apr 23 2015 *)

PROG

(Sage) [lucas_number1(n, 21, 1) for n in xrange(0, 20)] # Zerinvary Lajos, Jun 25 2008

CROSSREFS

Cf. R_3=A001906, R_4=A001353, R_5=A004254, R_6=A001109, R_7=A004187, R_8=A001090, R_9=A018913, R_10=A004189, R_11=A004190, R_12=A004191, R_13=A078362, R_14=A007655, R_15=A078364, R_16=A077412, R_17=A078366, R_18=A049660, R_19=A078368, R_20=A075843, R_21=this, sequence, R_22=A077421. See also A041219 and A041917.

Sequence in context: A215856 A218840 A171326 * A207776 A207273 A207766

Adjacent sequences:  A092496 A092497 A092498 * A092500 A092501 A092502

KEYWORD

easy,nonn

AUTHOR

Rainer Rosenthal, Apr 05 2004

EXTENSIONS

Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

Corrected by T. D. Noe, Nov 07 2006

STATUS

approved

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Last modified October 24 06:43 EDT 2017. Contains 293836 sequences.