A097778 Chebyshev polynomials S(n,23) with Diophantine property.

1, 23, 528, 12121, 278255, 6387744, 146639857, 3366328967, 77278926384, 1774048977865, 40725847564511, 934920445005888, 21462444387570913, 492701300469125111, 11310667466402306640, 259652650426783927609

Offset: 0

Author

Wolfdieter Lang, Aug 31 2004

Keywords

• nonn
• easy

Comments

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 squarefree.

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 23'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,...,22}. - Milan Janjic, Jan 25 2015

Examples

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

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..733
• R. Flórez, R. A. Higuita, A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
• Tanya Khovanova, Recursive Sequences
• Index entries for sequences relate d to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (23,-1).

Crossrefs

A004254, A049310.

Programs

Mathematica

LinearRecurrence[{23,-1},{1,23},20] (* Harvey P. Dale, May 06 2016 *)

Sage

[lucas_number1(n,23,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

Formula

a(n) = S(n, 23) = U(n, 23/2) = S(2*n+1, 5)/5 with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 5) = A004254(n+1).

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).

a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*22^k. - Philippe Deléham, Feb 10 2012

Product {n >= 0} (1 + 1/a(n)) = 1/21*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012

Product {n >= 1} (1 - 1/a(n)) = 1/46*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012