0,3

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.

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

Zerinvary Lajos, Sage Notebooks

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

sage: [lucas_number1(n, 21, 1) for n in xrange(0, 20)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

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: A097833 A064108 A067895 this_sequence A009965 A041842 A076552

Adjacent sequences: A092496 A092497 A092498 this_sequence A092500 A092501 A092502

easy,nonn

Rainer Rosenthal (r.rosenthal(AT)web.de), Apr 05 2004

Extension, Chebyshev and Pell comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Aug 31 2004

Corrected by T. D. Noe (noe(AT)sspectra.com), Nov 07 2006