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A097781
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Chebyshev polynomials S(n,27) with Diophantine property.
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7
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1, 27, 728, 19629, 529255, 14270256, 384767657, 10374456483, 279725557384, 7542215592885, 203360095450511, 5483180361570912, 147842509666964113, 3986264580646460139, 107481301167787459640, 2898008866949614950141
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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 - 725*a(n)^2 = +4 together with b(n)=A090248(n+1), n>=0. Note that D=725=29*5^2 is not squarefree.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 27'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,...,26}. - Milan Janjic, Jan 26 2015
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LINKS
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FORMULA
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a(n) = S(n, 27) = U(n, 27/2) = S(2*n+1, sqrt(29))/sqrt(29) 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) = 27*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=27; a(-1)=0.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (27+5*sqrt(29))/2 and am = (27-5*sqrt(29))/2.
G.f.: 1/(1-27*x+x^2).
Product {n >= 0} (1 + 1/a(n)) = 1/5*(5 + sqrt(29)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 5/54*(5 + sqrt(29)). - Peter Bala, Dec 23 2012
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EXAMPLE
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(x,y) = (27;1), (727;27), (19602;728), ... give the positive integer solutions to x^2 - 29*(5*y)^2 =+4.
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MAPLE
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with (combinat):seq(fibonacci(2*n, 5)/5, n=1..16); # Zerinvary Lajos, Apr 20 2008
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MATHEMATICA
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CoefficientList[Series[1/(1 - 27 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 24 2012 *)
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PROG
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(Sage) [lucas_number1(n, 27, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
(Magma) I:=[1, 27, 728]; [n le 3 select I[n] else 27*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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