0,2

All positive integer solutions of Pell equation b(n)^2 - 525*a(n)^2 = +4 together with b(n)=A090731(n+1), n>=0. Note that D=525=21*5^2 is not square-free.

Tanya Khovanova, Recursive Sequences

Index entries for sequences relate d to Chebyshev polynomials.

a(n)= S(n, 23)=U(n, 23/2)= S(2*n+1, sqrt(25))/sqrt(25) 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)=23*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=23; a(-1)=0.

a(n)=(ap^(n+1) - am^(n+1))/(ap-am) with ap := (23+5*sqrt(21))/2 and am := (23-5*sqrt(21))/2.

G.f.: 1/(1-23*x+x^2).

(x,y) = (23;1), (527;23), (12098;528), ... give the positive integer solutions to x^2 - 21*(5*y)^2 =+4.

sage: [lucas_number1(n, 23, 1) for n in xrange(1, 20)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

Adjacent sequences: A097775 A097776 A097777 this_sequence A097779 A097780 A097781

Sequence in context: A136285 A114926 A118338 this_sequence A057193 A014960 A009967

nonn,easy

Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Aug 31 2004