OFFSET
0,2
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
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 linear recurrences with constant coefficients, signature (23,-1).
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
EXAMPLE
(x,y) = (23;1), (527;23), (12098;528), ... give the positive integer solutions to x^2 - 21*(5*y)^2 =+4.
MATHEMATICA
LinearRecurrence[{23, -1}, {1, 23}, 20] (* Harvey P. Dale, May 06 2016 *)
PROG
(Sage) [lucas_number1(n, 23, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 31 2004
STATUS
approved