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Chebyshev polynomials of the second kind, U(n,x), evaluated at x=15.
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%I #67 Jul 08 2023 04:03:41

%S 0,1,30,899,26940,807301,24192090,724955399,21724469880,651009141001,

%T 19508549760150,584605483663499,17518655960144820,524975073320681101,

%U 15731733543660288210,471427031236487965199,14127079203550978667760

%N Chebyshev polynomials of the second kind, U(n,x), evaluated at x=15.

%C b(n+1)^2 - 14*(4*a(n))^2 = +1, n>=-1, with b(n)=A068203(n) gives all nonnegative integer solutions of this D=224 Pell equation.

%C For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 30'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 About first comment, more generally, for t(m) = m + sqrt(m^2-1) and u(n) = (t(m)^(n+1) - 1/t(m)^(n+1))/(t(m) - 1/t(m)), we can verify that ((u(n+1) - u(n-1))/2)^2 - (m^2-1)*u(n)^2 = 1. In this case is m=15. - _Bruno Berselli_, Nov 21 2011

%C For n>=0, a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,29}. - _Milan Janjic_, Jan 26 2015

%H Vincenzo Librandi, <a href="/A097313/b097313.txt">Table of n, a(n) for n = -1..200</a>

%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 related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (30,-1).

%F a(n) = S(n, 30) = U(n, 15), n>=-1, with Chebyshev polynomials of 2nd kind. See A049310 for the triangle of S(n, x) coefficients. S(-1, x) := 0 =: U(-1, x).

%F G.f.: 1/(1-30*x+x^2).

%F a(n) = ((15+4*sqrt(14))^(n+1) - (15-4*sqrt(14))^(n+1))/(8*sqrt(14)) (Binet form).

%F a(n) = sqrt((A068203(n+1)^2 - 1)/224), n>=-1.

%F a(n) = 30*a(n-1)-a(n-2) for n>0; a(-1)=0, a(0)=1. - _Philippe Deléham_, Nov 18 2008

%F a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*29^k. - _Philippe Deléham_, Feb 10 2012

%F With an offset of 0, product {n >= 1} (1 + 1/a(n)) = 1/7*(7 + 2*sqrt(14)). - _Peter Bala_, Dec 23 2012

%F Product {n >= 2} (1 - 1/a(n)) = 1/15*(7 + 2*sqrt(14)). - _Peter Bala_, Dec 23 2012

%p 0,seq(orthopoly[U](n,15),n=0..50); # _Robert Israel_, Jan 26 2015

%t Table[GegenbauerC[n, 1, 15], {n,-1,20}] (* _Vladimir Joseph Stephan Orlovsky_, Sep 11 2008 *)

%t LinearRecurrence[{30,-1},{0,1},50] (* _Vincenzo Librandi_, Feb 12 2012 *)

%t ChebyshevU[Range[22] -2, 15] (* _G. C. Greubel_, Dec 22 2019 *)

%o (Sage) [lucas_number1(n,30,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 27 2008

%o (Sage) [chebyshev_U(n,15) for n in (-1..20)] # _G. C. Greubel_, Dec 22 2019

%o (Magma) I:=[0, 1]; [n le 2 select I[n] else 30*Self(n-1)-Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Feb 12 2012

%o (PARI) vector( 21, n, polchebyshev(n-2, 2, 15) ) \\ _G. C. Greubel_, Dec 22 2019

%o (GAP) m:=15;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Dec 22 2019

%Y Cf. A200442.

%Y Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), this sequence (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).

%K nonn,easy

%O -1,3

%A _Wolfdieter Lang_, Aug 31 2004