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A097780
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Chebyshev polynomials S(n,25) with Diophantine property.
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3
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1, 25, 624, 15575, 388751, 9703200, 242191249, 6045078025, 150884759376, 3766073906375, 94000962899999, 2346257998593600, 58562449001940001, 1461714967049906425, 36484311727245720624, 910646078214093109175
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OFFSET
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0,2
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COMMENTS
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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.
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
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
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LINKS
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Indranil Ghosh, Table of n, a(n) for n = 0..714
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (25,-1).
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FORMULA
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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).
a(n)=25*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=25; a(-1)=0.
a(n)=(ap^(n+1) - am^(n+1))/(ap-am) with ap := (25+3*sqrt(69))/2 and am := (25-3*sqrt(69))/2.
G.f.: 1/(1-25*x+x^2).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*24^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/23*(23 + 3*sqrt(69)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/50*(23 + 3*sqrt(69)). - Peter Bala, Dec 23 2012
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EXAMPLE
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(x,y) = (2,0), (25;1), (623;25), (15550;624), ... give the nonnegative integer solutions to x^2 - 69*(3*y)^2 =+4.
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PROG
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(Sage) [lucas_number1(n, 25, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
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CROSSREFS
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Sequence in context: A307145 A061614 A171330 * A209222 A207691 A207926
Adjacent sequences: A097777 A097778 A097779 * A097781 A097782 A097783
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KEYWORD
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nonn,easy,changed
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AUTHOR
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Wolfdieter Lang, Aug 31 2004
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STATUS
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approved
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