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A097780 Chebyshev polynomials S(n,25) with Diophantine property. 3

%I #42 Sep 02 2021 04:01:11

%S 1,25,624,15575,388751,9703200,242191249,6045078025,150884759376,

%T 3766073906375,94000962899999,2346257998593600,58562449001940001,

%U 1461714967049906425,36484311727245720624,910646078214093109175

%N Chebyshev polynomials S(n,25) with Diophantine property.

%C 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.

%C 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

%C 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

%H Indranil Ghosh, <a href="/A097780/b097780.txt">Table of n, a(n) for n = 0..714</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 (25,-1).

%F 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).

%F a(n) = 25*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=25; a(-1)=0.

%F a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (25+3*sqrt(69))/2 and am := (25-3*sqrt(69))/2.

%F G.f.: 1/(1-25*x+x^2).

%F a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*24^k. - _Philippe Deléham_, Feb 10 2012

%F Product {n >= 0} (1 + 1/a(n)) = 1/23*(23 + 3*sqrt(69)). - _Peter Bala_, Dec 23 2012

%F Product {n >= 1} (1 - 1/a(n)) = 1/50*(23 + 3*sqrt(69)). - _Peter Bala_, Dec 23 2012

%e (x,y) = (2,0), (25;1), (623;25), (15550;624), ... give the nonnegative integer solutions to x^2 - 69*(3*y)^2 =+4.

%t LinearRecurrence[{25,-1},{1,25},20] (* _Harvey P. Dale_, Aug 23 2021 *)

%o (Sage) [lucas_number1(n,25,1) for n in range(1,20)] # _Zerinvary Lajos_, Jun 25 2008

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Aug 31 2004

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Last modified March 19 01:57 EDT 2024. Contains 370952 sequences. (Running on oeis4.)