A097780 Chebyshev polynomials S(n,25) with Diophantine property.

1, 25, 624, 15575, 388751, 9703200, 242191249, 6045078025, 150884759376, 3766073906375, 94000962899999, 2346257998593600, 58562449001940001, 1461714967049906425, 36484311727245720624, 910646078214093109175

Offset: 0

Author

Wolfdieter Lang, Aug 31 2004

Keywords

• nonn
• easy

Comments

All positive integer solutions of Pell equation b(n)^2 - 621*a(n)^2 = +4 together with b(n)=A090733(n+1), n>=0. Note that D=621=69*3^2 is not squarefree.

For positive n, a(n) equals the permanent of the tridiagonal matrix with 25'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,...,24}. - Milan Janjic, Jan 25 2015

Examples

			(x,y) = (2,0), (25;1), (623;25), (15550;624), ... give the nonnegative integer solutions to x^2 - 69*(3*y)^2 =+4.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..714
• Tanya Khovanova, Recursive Sequences
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (25,-1).

Programs

Mathematica

LinearRecurrence[{25,-1},{1,25},20] (* Harvey P. Dale, Aug 23 2021 *)

Sage

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

Formula

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

a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (25+3*sqrt(69))/2 and am := (25-3*sqrt(69))/2.

G.f.: 1/(1-25*x+x^2).

a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*24^k. - Philippe Deléham, Feb 10 2012

Product {n >= 0} (1 + 1/a(n)) = 1/23*(23 + 3*sqrt(69)). - Peter Bala, Dec 23 2012

Product {n >= 1} (1 - 1/a(n)) = 1/50*(23 + 3*sqrt(69)). - Peter Bala, Dec 23 2012