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A097778
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Chebyshev polynomials S(n,23) with Diophantine property.
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6
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1, 23, 528, 12121, 278255, 6387744, 146639857, 3366328967, 77278926384, 1774048977865, 40725847564511, 934920445005888, 21462444387570913, 492701300469125111, 11310667466402306640, 259652650426783927609
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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 - 525*a(n)^2 = +4 together with b(n)=A090731(n+1), n>=0. Note that D=525=21*5^2 is not squarefree.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 23'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,...,22}. - Milan Janjic, Jan 25 2015
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LINKS
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FORMULA
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a(n) = S(n, 23) = U(n, 23/2) = S(2*n+1, 5)/5 with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 5) = A004254(n+1).
a(n) = 23*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=23; a(-1)=0.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (23+5*sqrt(21))/2 and am := (23-5*sqrt(21))/2.
G.f.: 1/(1-23*x+x^2).
Product {n >= 0} (1 + 1/a(n)) = 1/21*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/46*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012
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EXAMPLE
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(x,y) = (23;1), (527;23), (12098;528), ... give the positive integer solutions to x^2 - 21*(5*y)^2 =+4.
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MATHEMATICA
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LinearRecurrence[{23, -1}, {1, 23}, 20] (* Harvey P. Dale, May 06 2016 *)
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PROG
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(Sage) [lucas_number1(n, 23, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
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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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