|
|
A097780
|
|
Chebyshev polynomials S(n,25) with Diophantine property.
|
|
3
|
|
|
1, 25, 624, 15575, 388751, 9703200, 242191249, 6045078025, 150884759376, 3766073906375, 94000962899999, 2346257998593600, 58562449001940001, 1461714967049906425, 36484311727245720624, 910646078214093109175
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
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
|
|
LINKS
|
|
|
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).
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
|
|
EXAMPLE
|
(x,y) = (2,0), (25;1), (623;25), (15550;624), ... give the nonnegative integer solutions to x^2 - 69*(3*y)^2 =+4.
|
|
MATHEMATICA
|
LinearRecurrence[{25, -1}, {1, 25}, 20] (* Harvey P. Dale, Aug 23 2021 *)
|
|
PROG
|
(Sage) [lucas_number1(n, 25, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|