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%I #46 Apr 26 2023 17:37:45
%S 1,23,528,12121,278255,6387744,146639857,3366328967,77278926384,
%T 1774048977865,40725847564511,934920445005888,21462444387570913,
%U 492701300469125111,11310667466402306640,259652650426783927609
%N Chebyshev polynomials S(n,23) with Diophantine property.
%C 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.
%C 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
%C 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
%H Indranil Ghosh, <a href="/A097778/b097778.txt">Table of n, a(n) for n = 0..733</a>
%H R. Flórez, R. A. Higuita, A. Mukherjee, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mukherjee/mukh2.html">Alternating Sums in the Hosoya Polynomial Triangle</a>, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences relate d to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (23,-1).
%F 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).
%F a(n) = 23*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=23; a(-1)=0.
%F a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (23+5*sqrt(21))/2 and am := (23-5*sqrt(21))/2.
%F G.f.: 1/(1-23*x+x^2).
%F a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*22^k. - _Philippe Deléham_, Feb 10 2012
%F Product {n >= 0} (1 + 1/a(n)) = 1/21*(21 + 5*sqrt(21)). - _Peter Bala_, Dec 23 2012
%F Product {n >= 1} (1 - 1/a(n)) = 1/46*(21 + 5*sqrt(21)). - _Peter Bala_, Dec 23 2012
%e (x,y) = (23;1), (527;23), (12098;528), ... give the positive integer solutions to x^2 - 21*(5*y)^2 =+4.
%t LinearRecurrence[{23,-1},{1,23},20] (* _Harvey P. Dale_, May 06 2016 *)
%o (Sage) [lucas_number1(n,23,1) for n in range(1,20)] # _Zerinvary Lajos_, Jun 25 2008
%Y A004254, A049310.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 31 2004
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