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 A029548 Expansion of 1/(1-32*x+x^2). 4

%I

%S 1,32,1023,32704,1045505,33423456,1068505087,34158739328,

%T 1092011153409,34910198169760,1116034330278911,35678188370755392,

%U 1140585993533893633,36463073604713840864,1165677769357309014015

%N Expansion of 1/(1-32*x+x^2).

%C From _Bruno Berselli_, Nov 21 2011: (Start)

%C A Diophantine property of these numbers: ((a(n+1)-a(n-1))/2)^2 - 255*a(n)^2 = 1.

%C 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. (End)

%C a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,31}. - _Milan Janjic_, Jan 26 2015

%H Vincenzo Librandi, <a href="/A029548/b029548.txt">Table of n, a(n) for n = 0..600</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 (32,-1).

%F a(n) = 32*a(n-1) - a(n-2), a(-1)=0, a(0)=1.

%F a(n) = S(n, 32) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind. See A049310. - _Wolfdieter Lang_, Nov 29 2002

%F a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap=16+sqrt(255) and am=16-sqrt(255).

%F a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*32^(n-2*k).

%F a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*31^k. - _Philippe DelĂ©ham_, Feb 10 2012

%F Product {n >= 0} (1 + 1/a(n)) = 1/15*(15 + sqrt(255)). - _Peter Bala_, Dec 23 2012

%F Product {n >= 1} (1 - 1/a(n)) = 1/32*(15 + sqrt(255)). - _Peter Bala_, Dec 23 2012

%t lst={};Do[AppendTo[lst, GegenbauerC[n, 1, 16]], {n, 0, 8^2}];lst (* _Vladimir Joseph Stephan Orlovsky_, Sep 11 2008 *)

%t CoefficientList[Series[1/(1 - 32 x + x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Dec 24 2012 *)

%o (Sage) [lucas_number1(n,32,1) for n in xrange(1, 16)] # _Zerinvary Lajos_, Nov 07 2009

%o (MAGMA) I:=[1, 32, 1023]; [n le 3 select I[n] else 32*Self(n-1)-Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Dec 24 2012

%Y Cf. A200441, A200442.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

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